16.4: Treatment Modalities
Learning Objectives
- Distinguish between the various modalities of treatment
- Discuss benefits of group therapy
Once a person seeks treatment, whether voluntarily or involuntarily, he has an intake done to assess his clinical needs. An intake is the therapist’s first meeting with the client. The therapist gathers specific information to address the client’s immediate needs, such as the presenting problem, the client’s support system, and insurance status. The therapist informs the client about confidentiality, fees, and what to expect in treatment. Confidentiality means the therapist cannot disclose confidential communications to any third party unless mandated or permitted by law to do so. During the intake, the therapist and client will work together to discuss treatment goals. Then a treatment plan will be formulated, usually with specific measurable objectives. Also, the therapist and client will discuss how treatment success will be measured and the estimated length of treatment. There are several different modalities of treatment (See figure \(\PageIndex{1}\)): Individual therapy, family therapy, couples therapy, and group therapy are the most common.
Indvidual Therapy
In individual therapy , also known as individual psychotherapy or individual counseling, the client and clinician meet one-on-one (usually from \(45\) minutes to \(1\) hour). These meetings typically occur weekly or every other week, and sessions are conducted in a confidential and caring environment (See figure \(\PageIndex{2}\)). The clinician will work with clients to help them explore their feelings, work through life challenges, identify aspects of themselves and their lives that they wish to change, and set goals to help them work towards these changes. A client might see a clinician for only a few sessions, or the client may attend individual therapy sessions for a year or longer. The amount of time spent in therapy depends on the needs of the client as well as her personal goals.
Group Therapy
In group therapy , a clinician meets together with several clients with similar problems (See figure \(\PageIndex{3}\)). When children are placed in group therapy, it is particularly important to match clients for age and problems. One benefit of group therapy is that it can help decrease a client’s shame and isolation about a problem while offering needed support, both from the therapist and other members of the group (American Psychological Association, 2014). A nine-year-old sexual abuse victim, for example, may feel very embarrassed and ashamed. If he is placed in a group with other sexually abused boys, he will realize that he is not alone. A child struggling with poor social skills would likely benefit from a group with a specific curriculum to foster special skills. A woman suffering from post-partum depression could feel less guilty and more supported by being in a group with similar women.
Group therapy also has some specific limitations. Members of the group may be afraid to speak in front of other people because sharing secrets and problems with complete strangers can be stressful and overwhelming. There may be personality clashes and arguments among group members. There could also be concerns about confidentiality: Someone from the group might share what another participant said to people outside of the group.
Another benefit of group therapy is that members can confront each other about their patterns. For those with some types of problems, such as sexual abusers, group therapy is the recommended treatment. Group treatment for this population is considered to have several benefits:
Group treatment is more economical than individual, couples, or family therapy. Sexual abusers often feel more comfortable admitting and discussing their offenses in a treatment group where others are modeling openness. Clients often accept feedback about their behavior more willingly from other group members than from therapists. Finally, clients can practice social skills in group treatment settings. (McGrath, Cumming, Burchard, Zeoli, & Ellerby, 2009)
Groups that have a strong educational component are called psycho-educational groups. For example, a group for children whose parents have cancer might discuss in depth what cancer is, types of treatment for cancer, and the side effects of treatments, such as hair loss. Often, group therapy sessions with children take place in school. They are led by a school counselor, a school psychologist, or a school social worker. Groups might focus on test anxiety, social isolation, self-esteem, bullying, or school failure (Shechtman, 2002). Whether the group is held in school or in a clinician’s office, group therapy has been found to be effective with children facing numerous kinds of challenges (Shechtman, 2002).
During a group session, the entire group could reflect on an individual’s problem or difficulties, and others might disclose what they have done in that situation. When a clinician is facilitating a group, the focus is always on making sure that everyone benefits and participates in the group and that no one person is the focus of the entire session. Groups can be organized in various ways: some have an overarching theme or purpose, some are time-limited, some have open membership that allows people to come and go, and some are closed. Some groups are structured with planned activities and goals, while others are unstructured: There is no specific plan, and group members themselves decide how the group will spend its time and on what goals it will focus. This can become a complex and emotionally charged process, but it is also an opportunity for personal growth (Page & Berkow, 1994).
Couples Therapy
Couples therapy involves two people in an intimate relationship who are having difficulties and are trying to resolve them (See figure \(\PageIndex{4}\)). The couple may be dating, partnered, engaged, or married. The primary therapeutic orientation used in couples counseling is cognitive-behavioral therapy (Rathus & Sanderson, 1999). Couples meet with a therapist to discuss conflicts and/or aspects of their relationship that they want to change. The therapist helps them see how their individual backgrounds, beliefs, and actions are affecting their relationship. Often, a therapist tries to help the couple resolve these problems, as well as implement strategies that will lead to a healthier and happier relationship, such as how to listen, how to argue, and how to express feelings. However, sometimes, after working with a therapist, a couple will realize that they are too incompatible and will decide to separate. Some couples seek therapy to work out their problems, while others attend therapy to determine whether staying together is the best solution. Counseling couples in a high-conflict and volatile relationship can be difficult. In fact, psychologists Peter Pearson and Ellyn Bader, who founded the Couples Institute in Palo Alto, California, have compared the experience of the clinician in couples’ therapy to be like “piloting a helicopter in a hurricane” (Weil, 2012, para. 7).
Family Therapy
Family therapy is a special form of group therapy, consisting of one or more families. Although there are many theoretical orientations in family therapy, one of the most predominant is the systems approach. The family is viewed as an organized system, and each individual within the family is a contributing member who creates and maintains processes within the system that shape behavior (Minuchin, 1985). Each member of the family influences and is influenced by the others. The goal of this approach is to enhance the growth of each family member as well as that of the family as a whole.
Often, dysfunctional patterns of communication that develop between family members can lead to conflict. A family with this dynamic might wish to attend therapy together rather than individually. In many cases, one member of the family has problems that detrimentally affect everyone. For example, a mother’s depression, teen daughter’s eating disorder, or father’s alcohol dependence could affect all members of the family. The therapist would work with all members of the family to help them cope with the issue, and to encourage resolution and growth in the case of the individual family member with the problem.
With family therapy, the nuclear family (i.e., parents and children) or the nuclear family plus whoever lives in the household (e.g., grandparent) come into treatment. Family therapists work with the whole family unit to heal the family. There are several different types of family therapy. In structural family therapy , the therapist examines and discusses the boundaries and structure of the family: who makes the rules, who sleeps in the bed with whom, how decisions are made, and what are the boundaries within the family. In some families, the parents do not work together to make rules, or one parent may undermine the other, leading the children to act out. The therapist helps them resolve these issues and learn to communicate more effectively.
In strategic family therapy , the goal is to address specific problems within the family that can be dealt with in a relatively short amount of time. Typically, the therapist would guide what happens in the therapy session and design a detailed approach to resolving each member’s problem (Madanes, 1991).
Summary
There are several modalities of treatment: individual therapy, group therapy, couples therapy, and family therapy are the most common. In an individual therapy session, a client works one-on-one with a trained therapist. In group therapy, usually \(5-10\) people meet with a trained group therapist to discuss a common issue (e.g., divorce, grief, eating disorders, substance abuse, or anger management). Couples therapy involves two people in an intimate relationship who are having difficulties and are trying to resolve them. The couple may be dating, partnered, engaged, or married. The therapist helps them resolve their problems as well as implement strategies that will lead to a healthier and happier relationship. Family therapy is a special form of group therapy. The therapy group is made up of one or more families. The goal of this approach is to enhance the growth of each individual family member and the family as a whole.
Glossary
- confidentiality
- therapist cannot disclose confidential communications to any third party, unless mandated or permitted by law
- couples therapy
- two people in an intimate relationship, such as husband and wife, who are having difficulties and are trying to resolve them with therapy
- family therapy
- special form of group therapy consisting of one or more families
- group therapy
- treatment modality in which \(5-10\) people with the same issue or concern meet together with a trained clinician
- individual therapy
- treatment modality in which the client and clinician meet one-on-one
- intake
- therapist’s first meeting with the client in which the therapist gathers specific information to address the client’s immediate needs
- strategic family therapy
- therapist guides the therapy sessions and develops treatment plans for each family member for specific problems that can addressed in a short amount of time
- structural family therapy
- therapist examines and discusses with the family the boundaries and structure of the family: who makes the rules, who sleeps in the bed with whom, how decisions are made, and what are the boundaries within the family
Contributors and Attributions
-
Rose M. Spielman with many significant contributors. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the creative commons license and may not be reproduced without the prior and express written consent of Rice University. For questions regarding this license, please contact<EMAIL_ADDRESS>.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


2.9.1: Measures of Variance and Standard Deviation - Loss of Information - Optional Material
Learning Objectives
- Consider the loss of information with grouped data
- Discuss class approximations
- Develop methods to approximate the range, variance, and standard deviation from grouped data
Section \(2.9.1\) Excel File (contains all of the data sets for this section)
Dispersion Measures on Interval-Grouped Data with Loss Of Information
What if we have data grouped over intervals instead of discrete single value groups as previously in Section \(2.9?\) As with our measures of central tendency, we have lost some information about the specific data values and are only able to roughly estimate our dispersion measures of range, variance, and standard deviation measures. We examine common approximation methods below; however, it should be known that the methods shown below are not unique as other similar but different choices can be made in what values are used from the class intervals.
The following table is again the frequency/relative frequency table based on data given by Florence Nightingale in her text Notes on Nursing (downloaded here ). We will again assume that Ms. Nightingale collected the data in a way such that if, for example, someone was in their \(29^{th}\) year of age (such as \(29.875\) years old), the data was reported as a \(29.\) We will also take this information as a representation of population data in our following computation work.
|
Age Intervals
(years) |
Interval Notation
(years) |
Frequency | Relative Frequency |
|---|---|---|---|
| \( 20 - 30 \) | \([20,30)\) | \( 1,441 \) | \( \frac {1,441}{25,466} \approx 0.0566 = 5.66\% \) |
| \( 30 - 39 \) | \([30,40)\) | \( 2,477 \) | \( \frac {2,477}{25,466} \approx 0.0973 = 9.73\% \) |
| \( 40 - 49 \) | \([40,50)\) | \( 4,971 \) | \( 0.1952 = 19.52\% \) |
| \( 50 - 59 \) | \([50,60)\) | \( 7,438 \) | \( 0.2921 = 29.21\% \) |
| \( 60 - 69 \) | \([60,70)\) | \( 6,367 \) | \( 0.2500 = 25.00\% \) |
| \( 70 - 79 \) | \([70,80)\) | \( 2,314 \) | \( 0.0909 = 9.09\% \) |
| \( 80+ \) | \([80,\text{above})\) | \( 458 \) | \( 0.0180 = 1.80\% \) |
| Totals: | \( 25,466 \) | \( 1.0000 = 100 \% \) |
We recall that we do not know, for example, how many \(20\) year old nurses there were in the data set, nor do we know how many \(27\) year old there were. We only know that there were \(1,441\) nurses reporting ages of \(20-29.\) This clearly implies that we cannot know what the actual data values were in the original data set; we have lost specific information about the original data set.
We will similarly approximate our measures of dispersions based on this grouped data by use of the midpoint value of each interval as our best approximation single measure for all values within the interval. So, again we will assume that all \(1,441\) people in their twenties are \(25\) years old, the midpoint of that interval. This is a drastic assumption in some sense, but with the loss of information on specific age measures in each interval, this is a reasonable way to approximate our dispersion measures just as with our central tendency measures. We will also again use \(85\) as our value for the last class interval of \([80,\text{above})\) even though the midpoint value could be larger if more was known about the actual data. This is also pointing to why it is not a general best practice when building frequency tables of data to use "and above" or "and below" within the last and first class interval descriptions; doing so provides even greater loss of key information about the data set.
We might choose to estimate the range measure using the largest and smallest midpoint values. That is, we estimate the range to be approximately \( 85\) \(- 25\) \(= 60 \) years. We note that we consider this a very rough estimate and major decisions should not be based on this estimate. Likely the range measure is larger, but again due to the loss of information when the data was grouped, we can't know for sure. We also note that others might estimate the range from such grouped data differently (such as the highest class' upper limit value minus the lowest class' lower limit value.)
Now for the variance estimate. First, we also recall from the optional Section \(2.8.1,\) that we computed a mean estimate value of \( \mu\) \(\approx 54.2675 \) years old. Variance, as the average of squared deviations from the mean, then leads to our producing the squared deviations column and weighting of those squared deviations by the relative frequency measures (we choose to use the relative frequency versus frequency approach in our work). Again, we use the midpoint of each interval as the data value to estimate deviation from the mean measures, and complete our work as in the non-interval grouped data approach of Section \(2.9.\)
Table \(\PageIndex{2}\): Computation of variance
|
Age Intervals
(years) |
Midpoint \(\left( m_j \right) \)
(years) |
\(P \left(m_j \right)\) | \( \left(m_j - \mu \right)^{2}\cdot P \left(m_j \right) \) |
|---|---|---|---|
| \( 20 - 29 \) | \( 25 \) | \( 0.0566 = 5.66\% \) | \( \left( 25 - 54.2675 \right)^{2} \cdot 0.0566 \approx 48.4828 \) |
| \( 30 - 39 \) | \( 35 \) | \( 0.0973 = 9.73\% \) | \( \left( 35 - 54.2675 \right)^{2} \cdot 0.0973 \approx 36.1213 \) |
| \( 40 - 49 \) | \( 45 \) | \( 0.1952 = 19.52\% \) | \( \left( 45 - 54.2675 \right)^{2} \cdot 0.1952 \approx 16.7651 \) |
| \( 50 - 59 \) | \( 55 \) | \( 0.2921 = 29.21\% \) | \( 0.1567 \) |
| \( 60 - 69 \) | \( 65 \) | \( 0.2500 = 25.00\% \) | \( 28.7966 \) |
| \( 70 - 79 \) | \( 75 \) | \( 0.0909 = 9.09\% \) | \( 39.0721 \) |
| \( 80+ \) | \( 85 \) | \( 0.0180 = 1.80\% \) | \( 17.0008 \) |
| Totals: | \( 1.0000 = 100 \% \) | \( \sum \left[ \left(m_j - \mu \right)^{2}\cdot P \left(m_j \right) \right] \approx 186.3954 \) |
So, we would estimate the variance of all this given population of non-domestic servant nurses in Great Britain to be about \( \sigma^{2}\) \(\approx 186.3954 \) years\(^{2} ,\) and hence the standard deviation to be about \( \sigma\) \(\approx \sqrt{186.3954}\) \(\approx 13.6527\) years. So, with interval-grouped data, we can estimate the variance and standard deviation by the same overall process, described symbolically by the given formulas with the use of each interval's midpoint represented by \( m_j:\)
Variance from an Interval-Grouped Distribution
\[ s^{2} \approx \frac {\sum \left[ \left( m_j - \bar{x} \right)^{2} \cdot f_j \right]}{\sum f_j - 1} \text{ ; when working with frequency interval-grouped sample data}\nonumber\]
\[ \sigma^{2} \approx \frac {\sum \left[ \left( m_j - \mu \right)^{2} \cdot f_j \right]}{\sum f_j} = \sum \left( m_j\cdot P(x_j) \right) \text{ when working with frequency or relative frequency interval-grouped population data}\nonumber\]
Standard Deviation from an Interval-Grouped Distribution
\[ s \approx \sqrt{s^{2}}\text{ when working with interval-grouped sample data} \nonumber\]
\[ \sigma^{2} \approx \sqrt{\sigma^{2}} \text{ when working with interval-grouped sample data} \nonumber\]
A bakery has been keeping records on the shelf-life of its best selling cinnamon rolls package. The bakery has sent the following frequency table asking for the median and mean measures of the data. Assuming this is sample data, find reasonable estimates of the range, variance, and standard deviation values of the data.
Table \(\PageIndex{3}\): Grouped frequency distribution for shelf-life data
|
Shelf-life
(days) |
Frequency |
|---|---|
| \( [3 , 8) \) | \(3\) |
| \( [8 , 13) \) | \(19\) |
| \( [13 , 18) \) | \(43\) |
| \( [18 , 23) \) | \(21\) |
| \( [23 , 28) \) | \(16\) |
| \( [28 , 33) \) | \(2\) |
- Answer
-
We again proceed by extending our table to include a column of midpoint values. Since this is sample data, we keep with frequency versus relative frequency measures in order to not lose sample size information.
Table \(\PageIndex{4}\): Preparatory computations using data from Table \(\PageIndex{3}\)
Shelf-life
(days)Midpoint \(\left( m_j \right) \)
(days)Frequency
\(f_j\)\( [3 , 8) \) \(\frac{3+8}{2}=5.5\) \(3\) \( [8 , 13) \) \(\frac{8+13}{2}=10.5\) \(19\) \( [13, 18)\) \(15.5\) \(43\) \( [18 , 23) \) \(20.5\) \(21\) \( [23, 28) \) \(25.5\) \(16\) \( [28 , 33) \) \(30.5\) \(2\) Totals: \(104\) First, we estimate the range to be \( 30.5- 5.5\) \(= 25\) days using our midpoint values. (As mentioned in the discussion above, one might instead choose to compute \(33 - 3\) \(=30\) days for the range; this estimate would be considered a maximum amount the range might truly be.)
Next, we estimate the variance. In Section \(2.8.1,\) we estimated the mean of this data to be \(17.1346\) days. So, to estimate the sample variance, we must form the weighted squared variations from the mean column, then sum those squared variations, and finally divide by one less than the sample size to form our "average" of the squared variations for sample variance purposes.
Table \(\PageIndex{5}\): Computation of variance
Shelf-life
(days)Midpoint \(\left( m_j \right) \)
(days)\( f_j \) \( \left(m_j - \mu \right)^{2}\cdot f_j\) \( [3 , 8) \) \(5.5\) \(3\) \( \left( 5.5 - 17.1346 \right)^{2} \cdot 3 \approx 406.0918 \) \( [8 , 13) \) \(10.5\) \(19\) \(\left( 10.5 - 17.1346 \right)^{2} \cdot 19 \approx 836.3404\) \( [13 , 18) \) \(15.5\) \(43\) \(\left( 15.5 - 17.1346 \right)^{2} \cdot 43 \approx 114.8924\) \( [18 , 23) \) \(20.5\) \(21\) \(237.8443\) \( [23 , 28) \) \(25.5\) \(16\) \(1119.6787\) \( [28 , 33) \) \(30.5\) \(2\) \(357.2678\) Totals: \(104\) \( s^{2} \approx \frac{\sum \left[ \left(m_j - \mu \right)^{2}\cdot f_j \right]}{\sum{\left(f_j\right)} -1} \approx \frac{3072.1154}{104 - 1}\approx 29.8264\) So, our estimate for the variance on the shelf-life of the packages of cinnamon rolls by this bakery would be about \(s^{2}\) \(\approx 29.8264\) days\(^{2}.\) And thus our standard deviation estimate would be \( s\) \(\approx \sqrt{29.8264}\) \(\approx 5.4614\) days. So the cinnamon roll packages roughly tend to last about \(17.1 \pm 5.5\) days.
In summary, we have now seen how we can produce rough estimates for the dispersion measurements when given interval-grouped data. We note how we are really just extending previous ideas/computations. However, we also remind ourselves that the resulting values should be used with caution in the interpretation of the dispersion of the data, not as if the values were the actual true measures of the data.


10.4.4: All The Slave Texts Thus Far
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We’ve been talking about slavery for a while now in class, inevitably.
Most recently, in Uncle Tom’s Cabin , the differences were interesting. Within the cast of black characters that we meet, I was struck by the amount “air time” they got compared to other slave texts. Eliza and George were the characters that we were following most of the time and thanks to the narration, we got to see a lot of what they were thinking. In Benito Cereno , it seemed like there was very little black perspective and in The Heroic Slave there was some but it really seemed to be the Mr. Listwell show. I think having the “air time”, so to speak, is important in gaining sympathy for characters. It’s easier to justify something as an anti-slavery text when there are legitimate perspectives form slaves.
The white characters in Uncle Tom’s Cabin also proved to have an interesting dichotomy, but depictions not too far off from the other books. There always seems to be a “good” white guy and a “bad” white guy in each of the texts (I use quotation marks because… well… it’s still slavery). In this book, so far, Mr. Shelby seems to embody our “good” white guy, or at least comparatively speaking. Mr. Listwell obviously was the good guy in The Heroic Slave . Then I think we were supposed to sympathize with Benito, even though I didn’t really like anyone in that book.
All of the parallels and differences between these slave texts give me the impression that Uncle Tom’s Cabin , and its fair distribution of “air time”, is meant to be a more progressive text for its time, even as it was published a few years prior to Benito Cereno and the same year as The Heroic Slave .


11.E: Exercises
Exercise \(\PageIndex{1}\)
Graphically, find the point \(\left( x_{1},y_{1}\right)\) which lies on both lines, \(x+3y=1\) and \(4x-y=3.\) That is, graph each line and see where they intersect.
- Answer
-
\(\begin{array}{c} x+3y=1 \\ 4x-y=3 \end{array}\), Solution is: \(\left[ x=\frac{10}{13},y=\frac{1}{13}\right]\).
Exercise \(\PageIndex{2}\)
Graphically, find the point of intersection of the two lines \(3x+y=3\) and \(x+2y=1.\) That is, graph each line and see where they intersect.
- Answer
-
\(\begin{array}{c} 3x+y=3 \\ x+2y=1 \end{array}\), Solution is: \(\left[ x=1,y=0\right]\)
Exercise \(\PageIndex{3}\)
You have a system of \(k\) equations in two variables, \(k\geq 2\). Explain the geometric significance of
- No solution.
- A unique solution.
- An infinite number of solutions.
Consider the following augmented matrix in which \(\ast\) denotes an arbitrary number and \(\blacksquare\) denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? \[\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & \blacksquare & \ast & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]\]
- Answer
-
The solution exists but is not unique.
Exercise \(\PageIndex{5}\)
Consider the following augmented matrix in which \(\ast\) denotes an arbitrary number and \(\blacksquare\) denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? \[\left[ \begin{array}{ccc|c} \blacksquare & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast \\ 0 & 0 & \blacksquare & \ast \end{array} \right]\]
- Answer
-
A solution exists and is unique.
Exercise \(\PageIndex{6}\)
Consider the following augmented matrix in which \(\ast\) denotes an arbitrary number and \(\blacksquare\) denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? \[\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & 0 & \ast & 0 & \ast \\ 0 & 0 & 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]\]
Exercise \(\PageIndex{7}\)
Consider the following augmented matrix in which \(\ast\) denotes an arbitrary number and \(\blacksquare\) denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? \[\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & 0 \\ 0 & 0 & 0 & 0 & \ast & \blacksquare \end{array} \right]\]
- Answer
-
There might be a solution. If so, there are infinitely many.
Exercise \(\PageIndex{8}\)
Suppose a system of equations has fewer equations than variables. Will such a system necessarily be consistent? If so, explain why and if not, give an example which is not consistent.
- Answer
-
No. Consider \(x+y+z=2\) and \(x+y+z=1.\)
Exercise \(\PageIndex{9}\)
If a system of equations has more equations than variables, can it have a solution? If so, give an example and if not, tell why not.
- Answer
-
These can have a solution. For example, \(x+y=1,2x+2y=2,3x+3y=3\) even has an infinite set of solutions.
Exercise \(\PageIndex{10}\)
Find \(h\) such that \[\left[ \begin{array}{rr|r} 2 & h & 4 \\ 3 & 6 & 7 \end{array} \right]\] is the augmented matrix of an inconsistent system.
- Answer
-
\(h=4\)
Exercise \(\PageIndex{11}\)
Find \(h\) such that \[\left[ \begin{array}{rr|r} 1 & h & 3 \\ 2 & 4 & 6 \end{array} \right]\] is the augmented matrix of a consistent system.
- Answer
-
Any \(h\) will work.
Exercise \(\PageIndex{12}\)
Find \(h\) such that \[\left[ \begin{array}{rr|r} 1 & 1 & 4 \\ 3 & h & 12 \end{array} \right]\] is the augmented matrix of a consistent system.
- Answer
-
Any \(h\) will work.
Exercise \(\PageIndex{13}\)
Choose \(h\) and \(k\) such that the augmented matrix shown has each of the following:
- one solution
- no solution
- infinitely many solutions
\[\left[ \begin{array}{rr|r} 1 & h & 2 \\ 2 & 4 & k \end{array} \right]\]
- Answer
-
If \(h\neq 2\) there will be a unique solution for any \(k\). If \(h=2\) and \( k\neq 4,\) there are no solutions. If \(h=2\) and \(k=4,\) then there are infinitely many solutions.
Exercise \(\PageIndex{14}\)
Choose \(h\) and \(k\) such that the augmented matrix shown has each of the following:
- one solution
- no solution
- infinitely many solutions
\[\left[ \begin{array}{rr|r} 1 & 2 & 2 \\ 2 & h & k \end{array} \right]\]
- Answer
-
If \(h\neq 4,\) then there is exactly one solution. If \(h=4\) and \(k\neq 4,\) then there are no solutions. If \(h=4\) and \(k=4,\) then there are infinitely many solutions.
Exercise \(\PageIndex{15}\)
Determine if the system is consistent. If so, is the solution unique? \[\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=1 \\ 2x+y-z=1 \\ 4x+2y+z=5 \end{array}\]
- Answer
-
There is no solution. The system is inconsistent. You can see this from the augmented matrix. \(\left[ \begin{array}{rrrrr} 1 & 2 & 1 & -1 & 2 \\ 1 & -1 & 1 & 1 & 1 \\ 2 & 1 & -1 & 0 & 1 \\ 4 & 2 & 1 & 0 & 5 \end{array} \right]\), : \(\left[ \begin{array}{rrrrr} 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & 0 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right] .\)
Exercise \(\PageIndex{16}\)
Determine if the system is consistent. If so, is the solution unique? \[\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=0 \\ 2x+y-z=1 \\ 4x+2y+z=3 \end{array}\]
- Answer
-
Solution is: \(\left[ w=\frac{3}{2}y-1,x=\frac{2}{3}-\frac{1}{2}y,z=\frac{1}{3 }\right]\)
Exercise \(\PageIndex{17}\)
Determine which matrices are in reduced row-echelon form.
- \(\left[ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 7 \end{array} \right]\)
- \(\left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right]\)
- \(\left[ \begin{array}{rrrrrr} 1 & 1 & 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 2 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 3 \end{array} \right]\)
- Answer
-
- This one is not.
- This one is.
- This one is.
Exercise \(\PageIndex{18}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 2 & -1 & 3 & -1 \\ 1 & 0 & 2 & 1 \\ 1 & -1 & 1 & -2 \end{array} \right]\]
Exercise \(\PageIndex{19}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & -1 \end{array} \right]\]
Exercise \(\PageIndex{20}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 3 & -6 & -7 & -8 \\ 1 & -2 & -2 & -2 \\ 1 & -2 & -3 & -4 \end{array} \right]\]
Exercise \(\PageIndex{21}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 2 & 4 & 5 & 15 \\ 1 & 2 & 3 & 9 \\ 1 & 2 & 2 & 6 \end{array} \right]\]
Exercise \(\PageIndex{22}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 4 & -1 & 7 & 10 \\ 1 & 0 & 3 & 3 \\ 1 & -1 & -2 & 1 \end{array} \right]\]
Exercise \(\PageIndex{23}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} 3 & 5 & -4 & 2 \\ 1 & 2 & -1 & 1 \\ 1 & 1 & -2 & 0 \end{array} \right]\]
Exercise \(\PageIndex{24}\)
Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. \[\left[ \begin{array}{rrrr} -2 & 3 & -8 & 7 \\ 1 & -2 & 5 & -5 \\ 1 & -3 & 7 & -8 \end{array} \right]\]
Exercise \(\PageIndex{25}\)
Find the solution of the system whose augmented matrix is \[\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 1 & 3 & 4 & 2 \\ 1 & 0 & 2 & 1 \end{array} \right]\]
Exercise \(\PageIndex{26}\)
Find the solution of the system whose augmented matrix is \[\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 2 & 0 & 1 & 1 \\ 3 & 2 & 1 & 3 \end{array} \right]\]
- Answer
-
The reduced row-echelon form is \(\left[ \begin{array}{rrr|r} 1 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{1}{4} & \frac{3}{4} \\ 0 & 0 & 0 & 0 \end{array} \right] .\) Therefore, the solution is of the form \(z=t,y=\frac{3}{4}+t\left( \frac{1}{4}\right) ,x=\frac{1}{2}-\frac{1}{2}t\) where \(t\in \mathbb{R}\).
Exercise \(\PageIndex{27}\)
Find the solution of the system whose augmented matrix is \[\left[ \begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 1 & 0 & 4 & 2 \end{array} \right]\]
- Answer
-
The reduced row-echelon form is \(\left[ \begin{array}{rrr|r} 1 & 0 & 4 & 2 \\ 0 & 1 & -4 & -1 \end{array} \right]\) and so the solution is \(z=t,y=-1+4t,x=2-4t.\)
Exercise \(\PageIndex{28}\)
Find the solution of the system whose augmented matrix is \[\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 1 & 2 & 0 & 0 & 1 & 3 \\ 1 & 0 & 1 & 0 & 2 & 2 \end{array} \right]\]
- Answer
-
The reduced row-echelon form is \(\left[ \begin{array}{rrrrr|r} 1 & 0 & 0 & 0 & 9 & 3 \\ 0 & 1 & 0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0 & -7 & -1 \\ 0 & 0 & 0 & 1 & 6 & 1 \end{array} \right]\) and so \(x_{5}=t,x_{4}=1-6t,x_{3}=-1+7t,x_{2}=4t,x_{1}=3-9t\).
Exercise \(\PageIndex{29}\)
Find the solution of the system whose augmented matrix is \[\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 0 & 2 & 0 & 0 & 1 & 3 \\ 1 & -1 & 2 & 2 & 2 & 0 \end{array} \right]\]
- Answer
-
The reduced row-echelon form is \(\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 0 & - \frac{1}{2} & \frac{5}{2} \\ 0 & 1 & 0 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]\). Therefore, let \(x_{5}=t,x_{3}=s.\) Then the other variables are given by \(x_{4}=-\frac{1}{2}-\frac{3}{2}t,x_{2}=\frac{3}{2}-t\frac{1}{2} ,,x_{1}=\frac{5}{2}+\frac{1}{2}t-2s.\)
Exercise \(\PageIndex{30}\)
Find the solution to the system of equations, \(7x+14y+15z=22,\) \(2x+4y+3z=5,\) and \(3x+6y+10z=13.\)
- Answer
-
Solution is: \(\left[ x=1-2t,z=1,y=t\right]\)
Exercise \(\PageIndex{31}\)
Find the solution to the system of equations, \(3x-y+4z=6,\) \(y+8z=0,\) and \(-2x+y=-4.\)
- Answer
-
Solution is: \(\left[ x=2-4t,y=-8t,z=t\right]\)
Exercise \(\PageIndex{32}\)
Find the solution to the system of equations, \(9x-2y+4z=-17,\) \(13x-3y+6z=-25,\) and \(-2x-z=3.\)
- Answer
-
Solution is: \(\left[x=-1,y=2,z=-1\right]\)
Exercise \(\PageIndex{33}\)
Find the solution to the system of equations, \(65x+84y+16z=546,\) \(81x+105y+20z=682,\) and \(84x+110y+21z=713.\)
- Answer
-
Solution is: \(\left[ x=2,y=4,z=5\right]\)
Exercise \(\PageIndex{34}\)
Find the solution to the system of equations, \(8x+2y+3z=-3,8x+3y+3z=-1,\) and \(4x+y+3z=-9.\)
- Answer
-
Solution is: \(\left[ x=1,y=2,z=-5\right]\)
Exercise \(\PageIndex{35}\)
Find the solution to the system of equations, \(-8x+2y+5z=18,-8x+3y+5z=13,\) and \(-4x+y+5z=19.\)
- Answer
-
Solution is: \(\left[x=-1,y=-5,z=4\right]\)
Exercise \(\PageIndex{36}\)
Find the solution to the system of equations, \(3x-y-2z=3,\) \(y-4z=0,\) and \(-2x+y=-2.\)
- Answer
-
Solution is: \(\left[ x=2t+1,y=4t,z=t\right]\)
Exercise \(\PageIndex{37}\)
Find the solution to the system of equations, \(-9x+15y=66,-11x+18y=79\), \(-x+y=4\), and \(z=3\).
- Answer
-
Solution is: \(\left[x=1,y=5,z=3\right]\)
Exercise \(\PageIndex{38}\)
Find the solution to the system of equations, \(-19x+8y=-108,\) \(-71x+30y=-404,\) \(-2x+y=-12,\) \(4x+z=14.\)
- Answer
-
Solution is: \(\left[ x=4,y=-4,z=-2\right]\)
Exercise \(\PageIndex{39}\)
Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain.
- Answer
-
No. Consider the system of equations \(x+y+2z=2\) and \(x+y+z=1\), this system has infinitely many solutions. If there is one, there will be a free variable because there are more variables than equations, and so there will be infinitely many solutions.
Exercise \(\PageIndex{40}\)
Suppose a system of linear equations has a \(2\times 4\) augmented matrix and the last column is a pivot column. Could the system of linear equations be consistent? Explain.
- Answer
-
No. This would lead to \(0=1.\)
Exercise \(\PageIndex{41}\)
Suppose the coefficient matrix of a system of \(n\) equations with \(n\) variables has the property that every column is a pivot column. Does it follow that the system of equations must have a solution? If so, must the solution be unique? Explain.
- Answer
-
Yes. It has a unique solution.
Exercise \(\PageIndex{42}\)
Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix?
- Answer
-
The last column must not be a pivot column. The remaining columns must each be pivot columns.
Exercise \(\PageIndex{43}\)
The steady state temperature, \(u\), of a plate solves Laplace’s equation, \(\Delta u=0.\) One way to approximate the solution is to divide the plate into a square mesh and require the temperature at each node to equal the average of the temperature at the four adjacent nodes. In the following picture, the numbers represent the observed temperature at the indicated nodes. Find the temperature at the interior nodes, indicated by \(x,y,z,\) and \(w\). One of the equations is \(z=\vspace{0.05in}\frac{1}{4}\left( 10+0+w+x\right)\).
(1,60) (100,0)
(1,1) (0,30) (1,0) 90 (0,60) (1,0) 90 (30,0) (0,1) 90 (60,0) (0,1) 90 (60,0) (30,0) (0,60) (0,30) (30,30) (30,60) (60,30) (60,60) (60,90) (30,90) (90,30) (90,60) (63,-4) \(10\) (33,-4) \(10\) (-10,60) \(20\) (-10,30) \(20\) (34,34) \(x\) (34,64) \(y\) (64,34) \(z\) (64,64) \(w\) (64,90) \(30\) (34,90) \(30\) (94,30) \(0\) (94,60) \(0\)
- Answer
-
You need \(\begin{array}{c} \frac{1}{4}\left( 20+30+w+x\right) -y=0 \\ \frac{1}{4}\left( y+30+0+z\right) -w=0 \\ \frac{1}{4}\left( 20+y+z+10\right) -x=0 \\ \frac{1}{4}\left( x+w+0+10\right) -z=0 \end{array}\), Solution is: \(\left[ w=15,x=15,y=20,z=10\right] .\)
Exercise \(\PageIndex{44}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrr} 4 & -16 & -1 & -5 \\ 1 & -4 & 0 & -1 \\ 1 & -4 & -1 & -2 \end{array} \right]\]
Exercise \(\PageIndex{45}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrr} 3 & 6 & 5 & 12 \\ 1 & 2 & 2 & 5 \\ 1 & 2 & 1 & 2 \end{array} \right]\]
Exercise \(\PageIndex{46}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 3 \\ 1 & 4 & 1 & 0 & -8 \\ 1 & 4 & 0 & 1 & 2 \\ -1 & -4 & 0 & -1 & -2 \end{array} \right]\]
Exercise \(\PageIndex{47}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrr} 4 & -4 & 3 & -9 \\ 1 & -1 & 1 & -2 \\ 1 & -1 & 0 & -3 \end{array} \right]\]
Exercise \(\PageIndex{48}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} 2 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 7 \\ 1 & 0 & 0 & 1 & 7 \end{array} \right]\]
Exercise \(\PageIndex{49}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrr} 4 & 15 & 29 \\ 1 & 4 & 8 \\ 1 & 3 & 5 \\ 3 & 9 & 15 \end{array} \right]\]
Exercise \(\PageIndex{50}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 1 \\ 1 & 2 & 3 & -2 & -18 \\ 1 & 2 & 2 & -1 & -11 \\ -1 & -2 & -2 & 1 & 11 \end{array} \right]\]
Exercise \(\PageIndex{51}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} 1 & -2 & 0 & 3 & 11 \\ 1 & -2 & 0 & 4 & 15 \\ 1 & -2 & 0 & 3 & 11 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]\]
Exercise \(\PageIndex{52}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrr} -2 & -3 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ -3 & 0 & -3 \end{array} \right]\]
Exercise \(\PageIndex{53}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} 4 & 4 & 20 & -1 & 17 \\ 1 & 1 & 5 & 0 & 5 \\ 1 & 1 & 5 & -1 & 2 \\ 3 & 3 & 15 & -3 & 6 \end{array} \right]\]
Exercise \(\PageIndex{54}\)
Find the rank of the following matrix. \[\left[ \begin{array}{rrrrr} -1 & 3 & 4 & -3 & 8 \\ 1 & -3 & -4 & 2 & -5 \\ 1 & -3 & -4 & 1 & -2 \\ -2 & 6 & 8 & -2 & 4 \end{array} \right]\]
Exercise \(\PageIndex{55}\)
Suppose \(A\) is an \(m\times n\) matrix. Explain why the rank of \(A\) is always no larger than \(\min \left( m,n\right) .\)
- Answer
-
It is because you cannot have more than \(\min \left( m,n\right)\) nonzero rows in the reduced row-echelon form. Recall that the number of pivot columns is the same as the number of nonzero rows from the description of this reduced row-echelon form.
Exercise \(\PageIndex{56}\)
State whether each of the following sets of data are possible for the matrix equation \(AX=B\). If possible, describe the solution set. That is, tell whether there exists a unique solution, no solution or infinitely many solutions. Here, \(\left[ A |B \right]\) denotes the augmented matrix.
- \(A\) is a \(5\times 6\) matrix, \(\mathrm{rank}\left( A\right) =4\) and \(\mathrm{rank}\left[ A|B \right] =4.\)
- \(A\) is a \(3\times 4\) matrix, \(\mathrm{rank}\left( A\right) =3\) and \(\mathrm{rank}\left[ A|B\right] =2.\)
- \(A\) is a \(4\times 2\) matrix, \(\mathrm{rank}\left( A\right) =4\) and \(\mathrm{rank}\left[ A|B \right] =4.\)
- \(A\) is a \(5\times 5\) matrix, \(\mathrm{rank}\left( A\right) =4\) and \(\mathrm{rank}\left[ A|B \right] =5.\)
- \(A\) is a \(4\times 2\) matrix, \(\mathrm{rank}\left( A\right) =2\) and \(\mathrm{rank}\left[ A|B \right] =2\).
- Answer
-
This says \(B\) is in the span of four of the columns. Thus the columns are not independent. Infinite solution set.
This surely can’t happen. If you add in another column, the rank does not get smaller.
This says \(B\) is in the span of the columns and the columns must be independent. You can’t have the rank equal 4 if you only have two columns.
This says \(B\) is not in the span of the columns. In this case, there is no solution to the system of equations represented by the augmented matrix.
In this case, there is a unique solution since the columns of \(A\) are independent.
Exercise \(\PageIndex{57}\)
Consider the system \(-5x+2y-z=0\) and \(-5x-2y-z=0.\) Both equations equal zero and so \(-5x+2y-z=-5x-2y-z\) which is equivalent to \(y=0.\) Does it follow that \(x\) and \(z\) can equal anything? Notice that when \(x=1\), \(z=-4,\) and \(y=0\) are plugged in to the equations, the equations do not equal \(0\). Why?
- Answer
-
These are not legitimate row operations. They do not preserve the solution set of the system.
Exercise \(\PageIndex{58}\)
Balance the following chemical reactions.
- \(KNO_{3}+H_{2}CO_{3}\rightarrow K_{2}CO_{3}+HNO_{3}\)
- \(AgI+Na_{2}S\rightarrow Ag_{2}S+NaI\)
- \(Ba_{3}N_{2}+H_{2}O\rightarrow Ba\left( OH\right) _{2}+NH_{3}\)
- \(CaCl_{2}+Na_{3}PO_{4}\rightarrow Ca_{3}\left( PO_{4}\right) _{2}+NaCl\)
Exercise \(\PageIndex{59}\)
In the section on dimensionless variables it was observed that \(\rho V^{2}AB\) has the units of force. Describe a systematic way to obtain such combinations of the variables which will yield something which has the units of force.
Exercise \(\PageIndex{60}\)
Consider the following diagram of four circuits.
The current in amps in the four circuits is denoted by \(I_{1},I_{2},I_{3},I_{4}\) and it is understood that the motion is in the counter clockwise direction. If \(I_{k}\) ends up being negative, then it just means the current flows in the clockwise direction.
In the above diagram, the top left circuit should give the equation \[2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5\] For the circuit on the lower left, you should have \[4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10\] Write equations for each of the other two circuits and then give a solution to the resulting system of equations.
- Answer
-
The other two equations are \[\begin{aligned} 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2} &=&-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1} &=&0\end{aligned}\] Then the system is \[\begin{array}{c} 2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5 \\ 4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10 \\ 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2}=-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1}=0 \end{array}\] The solution is: \[\begin{aligned} I_{1}&=& -\frac{750}{373} \\ I_{2}&=& -\frac{1421}{1119} \\ I_{3}&=& -\frac{3061}{1119} \\ I_{4}&=& -\frac{1718}{1119}\end{aligned}\]
Exercise \(\PageIndex{61}\)
Consider the following diagram of three circuits.
The current in amps in the four circuits is denoted by \(I_{1},I_{2},I_{3}\) and it is understood that the motion is in the counter clockwise direction. If \(I_{k}\) ends up being negative, then it just means the current flows in the clockwise direction.
Find \(I_{1},I_{2},I_{3}\).
- Answer
-
You have \[\begin{aligned} 2I_{1}+5I_{1}+3I_{1}-5I_{2} &=& 10 \\ I_{2}- I_{3} +3I_{2}+7I_{2}+5I_{2}-5I_{1} &=&-12 \\ 2I_{3}+4I_{3}+4I_{3}+I_{3}-I_{2} &=& 0\end{aligned}\] Simplifying this yields \[\begin{aligned} 10I_{1}-5I_{2} &=& 10 \\ -5I_{1} + 16I_{2}- I_{3} &=&-12 \\ -I_{2} + 11I_{3} &=&0\end{aligned}\] The solution is given by \[I_{1}=\frac{218}{295},I_{2}=-\frac{154}{295},I_{3}=-\frac{14}{295}\]


7.10: Writing Conclusions
Studies have shown that the human brain is more likely to remember items at the beginning and end of lists, presentations, and other texts. Recollection of the last thing a person reads or hears is called the recency effect – they’re remembering the most recent information they’ve encountered. This is why your conclusion is so important. It’s your final chance to make an impression on your readers.
Conclusions have two jobs :
- Leave readers with something to think about.
- Clarify why your topic matters to them and the larger community. Some people refer to a conclusion as the so what ? section of a text, because it helps readers understand the significance of your subject. See Figure 7.7 .
Figure 7.7: The conclusion as the “so what?” section
What techniques can keep readers thinking about the topic at the end of a piece of writing? Some of the same methods that work for the introduction also work for the conclusion, but the formula is a little different:
You’re trying to end your piece, so don’t start making new claims or sharing new research. Instead, try to help readers see how they relate to your subject matter. Sometimes this means suggesting that the reader do something specific: a call to action. You could also end by raising questions related to your topic or by making suggestions for how this topic may develop in the future. Leaving readers with interesting ideas to think about is the key to a successful conclusion. See Film Clip 7.2 about writing conclusions.
Film Clip 7.2 : Writing conclusions [1:03]
See Table 7.6 for methods and examples for concluding your writing and giving your readers a sense of closure or an idea of what you would like them to think about or do next.
Table 7.6: Methods and examples for conclusions
| Method | Example |
|---|---|
|
Make a call to action: prompt readers to do something. |
Nurses are uniquely positioned to advocating for better public policy to limit and restrict the consumption of sugar. |
|
Ask a rhetorical question: a question asked for mere effect. A rhetorical question is meant to make people think, but not necessarily come to an answer. Often, the answer to rhetorical questions is clear right away, but the deeper significance needs to be pondered. |
Have you considered the impact of sugar consumption on your health or the clients you care for? |
Student Tip
Optimizing Your Conclusion
Avoid introducing new ideas in the conclusion. But don’t simply copy and paste your thesis from earlier in the text. You’ve likely had a whole conversation in the text since the reader first encountered that thesis. Simply repeating it, or even replacing a few key words with synonyms, doesn’t acknowledge that bigger conversation. Instead, try pointing the reader back to the main idea in a new way.
Attribution statement
Content from this page was remixed with our own original content and adapted, with editorial and organizational changes, from:
The Word on College Reading and Writing by Carol Burnell, Jaime Wood, Monique Babin, Susan Pesznecker, and Nicole Rosevear, licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.Download for free at: https://openoregon.pressbooks.pub/wrd/


6.3: Proteins
Amino acids
There are 20 naturally occurring amino acids. They have the general format shown below.
Credit: Amino Acid Structure by TechGuy78
These molecules are called amino acids because they all share two functional groups in common: an amine group (-NH2) and a carboxylic acid group (-COOH). With the exception of proline, they all have the same backbone: NH2-CH-COOH. They each have a different R- group or side chain attached to the first carbon atom.
The twenty naturally occurring amino acids (plus selenocysteine) are shown below. Can you identify the backbone in each image? The rest of the molecule is the side chain.
Credit: Amino Acids by Dan Cojocari
There is a wide variety of side chains and they give the amino acids different properties. Some are hydrophobic (from the Greek for water and fear) and cannot dissolve in water because they are nonpolar. Others are hydrophilic (from the Greek for water and love) and can dissolve in water. This affects protein folding (discussed below) because hydrophobic amino acids tend to be found on the inside of proteins whereas hydrophilic proteins tend to be found on the outer edges of proteins (near aqueous solutions).
The Peptide Bond
Amino acids are monomers that form proteins , naturally occurring polymers . The bond between amino acids is called a peptide bond. It forms between the carboxylic acid group of one amino acid and the amine group of another amino acid. Thus, the backbones of the amino acids get connected to one another and the side chains do not connect to one another. The resulting chain, or protein, looks like a charm bracelet, a long backbone with various different attachments at each amino acid. The general form of the reaction is:
amino acid 1 + amino acid 2 → dipeptide + water
This reaction is called condensation because water is produced. To see where the water comes from, examine the structures involved:
When the reaction occurs, the carboxylic acid group loses the OH and the amine group loses an H. Together, these form water. The peptide bond forms between the carbon of amino acid 1 and the nitrogen of amino acid 2:
When this reaction repeats, additional amino acids are added to the chain. Because water is removed, the backbone of a protein has the pattern N-C-C-N-C-C-N-C-C (there is an amine group on the first amino acid and a carboxylic acid group on the last amino acid). Therefore, you can determine how many amino acids are in a peptide by counting how many times the -N-C-C- pattern repeats.
Peptide sequences are always written with the N-terminus, the amine end, on the left and the C-terminus, the carboxylic acid end, on the right.
Protein Structure
Because proteins are very large molecules, we can think about their structure at four different levels. Sometimes we are interested in the order in which the amino acids are connected in the chain. This is the primary structure , 1°, of the protein.
-Trp-Tyr-Ser-Ala-Leu-
It lists the amino acids present in the correct order (from N-terminus to C-terminus), but tells nothing about the three-dimensional shape of the protein.
The next level of protein structure is the secondary structure , 2°. This indicates a local 3D structure, or in other words, interactions that are occurring in a portion of the protein. It describes the structure in only a portion of the protein chain. Therefore, a protein may contain many secondary structures. Some common secondary structures are the alpha helix (⍺-helix) and the beta-sheet (ꞵ-sheet). In these secondary structures hydrogen bonds hold different parts of the protein backbone together in a spiral (the alpha helix) or in strands that go back and forth (the beta sheet).
Credit: Pi Helix by WillowW
On the left above is an alpha helix where the pink lines indicate hydrogen bonds between N-H and C=O portions of the backbone. To the right is a portion of a beta-sheet where hydrogen bonds are indicated by dashed lines that connect two portions of the same protein (it folds over along itself). This pattern continues with the protein backbone folding again and again as shown using a ribbon diagram in the image on the right below.
Credit: Protein Image from OPM database by Andrei Lomize
The figure to the left above shows an entire protein. The backbone in portions of the structure is shown as a ribbon to emphasize the local, or secondary, structure: an alpha-helix. Other portions of the backbone are shown as tubes. Similarly, the figure on the right shows a beta-sheet with the portions of the backbone that are hydrogen bonding represented with ribbons.
Thus, you can see that the secondary structure does not indicate the overall shape of a protein, just the structure in a particular portion of the backbone. The tertiary structure (3°) does refer to the entire 3D shape of a protein strand. Because the tertiary structure of each protein is unique, a collaborative effort has been made to create a record of all known protein structures. The result is the Protein Data Bank .
It may seem as if no additional levels of structure would be needed. However, some proteins are composed of multiple strands interacting with one another. The overall arrangement of these strands is the quaternary structure (4°). Many proteins do not have a quaternary structure because they consist of only one chain. Hemoglobin is an example of a protein that does have a quaternary structure because it has four strands. Its quaternary structure can be seen below.
Credit: Structure of haemoglobin by Richard Wheeler.
The four strands that make up this protein are called subunits and are shown in red and blue in the picture.
Enzymes
Enzymes are proteins that act as catalysts. They speed up reactions that would otherwise take too long within the body. This is preferable to another method of speeding up reactions: increasing the temperature! Instead, mammals keep their body temperatures regulated and enzymes perform well in this range.
There are many different enzymes in the body because they each catalyze specific chemical reactions. Name of enzymes typically end with the suffix -ase and the rest of the name often provides a clue as to the function of the enzyme, e.g. the enzyme lactase breaks lactose down into two simple sugars and DNA polymerase is used in the synthesis of DNA (a polymer).
Enzymes provide a location for one or more reactants to bind while the reaction is occurring and lower the activation energy of the reaction. In these reactions the reactants are called substrates and the location on the enzyme where the reaction occurs is called the active site.
The picture below shows the interaction between substrates and an enzyme for a reaction that follows the induced-fit model. In the induced-fit model of enzyme catalysis, when the substrate(s) bind in the active site the protein changes structure slightly to fit around the substrate(s).
Credit: Hexokinase induced fit by Thomas Shafee
The top of the diagram is the “before” picture where the enzyme and two substrate molecules are initially separate from one another. In both the block diagram on the left and the molecular model on the right the active site is highlighted in blue and the substrate molecules are shown in black. In the bottom half of the diagram, the substrates are bound to the enzyme and the active site has closed around them. After the reaction is completed (not shown) the product will leave the active site and the enzyme will return to its original conformation to be used again by new substrate molecules. Thus enzymes, like all catalysts, are not consumed in chemical reactions and can be reused by other substrate molecules.
Another model of enzyme function is the lock-and-key model. This model is similar to the induced-fit model in that the substrate(s) bind to an active site in the enzyme, but there is no change in the structure of the enzyme. The substrate(s) fit into the active site like a key fits into a lock.
In order to control reactions occurring in the body, inhibitors can be produced. Inhibitors prevent enzymes from catalyzing chemical reactions, for instance when enough product has already been produced. There is a delicate balance between the amounts of substrates, products, and inhibitors so that the body functions properly.
Some inhibitors disrupt the chemical reactions by stealing the active site. These are called competitive inhibitors and they have a shape similar to the substrate so that they can fit into the active site. There are a limited number of enzyme molecules present and whenever an inhibitor molecule binds with one there is one less opportunity for the substrate to react. Increasing the concentration of inhibitor molecules slows down the reaction whereas increasing the concentration of substrate molecules overcomes the inhibition. Thus, as concentrations of substrates and inhibitors change, the amount of product being produced changes.
Other inhibitors do not bind to the active site, they have a totally different shape than the substrates that they inhibit.


12.2: Permutations and Combinations
Consider the following counting problems:
- In how many ways can three runners finish a race?
- In how many ways can a group of three people be chosen to work on a project?
What is the difference between these two problems? In the first problem the order that the runners finish the race matters. In the second problem the order in which the three people are chosen is not important, only which three people are chosen matters.
A permutation is an arrangement of a set of items. The number of permutations of n items taking r at a time is given by:
\[P(n, r)=\frac{n !}{(n-r) !} \label{permutation}\]
Note: Many calculators can calculate permutations directly. Look for a function that looks like \(_nP_r\) or \(P(n,r)\)
Let’s look at a simple example to understand the formula for the number of permutations of a set of objects. Assume that 10 cars are in a race. In how many ways can three cars finish in first, second and third place? The order in which the cars finish is important. Use the multiplication principle. There are 10 possible cars to finish first. Once a car has finished first, there are nine cars to finish second. After the second car is finished, any of the eight remaining cars can finish third. 10 x 9 x 8 = 720. This is a permutation of 10 items taking three at a time.
Using the permutation formula:
\[P(10,3)=\frac{10 !}{(10-3) !}=\frac{10 !}{7 !}= \dfrac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 10 \cdot 9 \cdot 8 = 720 \nonumber\]
Using the fundamental counting principle:
\[\underline{10} \cdot \underline{9} \cdot \underline{8} = 720 \nonumber\]
There are 720 different ways for cars to finish in the top three places.
The school orchestra is planning to play six pieces of music at their next concert. How many different programs are possible?
Solution
This is a permutation because they are arranging the songs in order to make the program. Using the permutation formula:
\[P(6,6)=\frac{6 !}{(6-6) !}=\frac{6 !}{0 !}=\frac{720}{1}=720 \nonumber\]
Using the fundamental counting principle:
\[\underline{6} \cdot \underline{5} \cdot \underline{4} \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=720 \nonumber\]
There are 720 different ways of arranging the songs to make the program.
The Volunteer Club has 18 members. An election is held to choose a president, vice-president and secretary. In how many ways can the three officers be chosen?
Solution
The order in which the officers are chosen matters so this is a permutation.
Using the permutation formula:
\[P(18,3)=\frac{18 !}{(18-3) !}=\frac{18 !}{15 !}=18 \cdot 17 \cdot 16=4896 \nonumber\]
Note: All digits in 18! in the numerator from 15 down to one will cancel with the 15! in the denominator.
Using the fundamental principle:
\(\begin{array} {cccccc} {\underline{18}}&{\cdot}&{\underline{17}}&{\cdot}&{\underline{16}}&{ = 4896}\\ {\text{Pres.}}&{}&{\text{V.P.}}&{}&{\text{Sec.}}&{} \end{array}\)
There are 4896 different ways the three officers can be chosen.
Another notation for permutations is \(_nP_r\). So, \(P(18,3)\) can also be written as \(_{18}P_3\). Most scientific calculators have an \(_nP_r\) button or function.
Combinations are when the order does not even matter. We are just collecting objects together.
Choose a committee of two people from persons A, B, C, D and E. By the multiplication principle there are \(5 \cdot 4 = 20\) ways to arrange the two people.
AB AC AD AE BA BC BD BE CA CB
CD CE DA DB DC DE EA EB EC ED
Committees AB and BA are the same committee. Similarly for committees CD and DC. Every committee is counted twice.
\[\frac{20}{2}=10 \nonumber\]
so there are 10 possible different committees.
Now choose a committee of three people from persons A, B, C, D and E. There are \(5 \cdot 4 \cdot 3 = 60\) ways to pick three people in order. Think about the committees with persons A, B and C. There are \(3! =6\) of them.
ABC ACB BAC BCA CAB CBA
Each of these is counted as one of the 60 possibilities but they are the same committee. Each committee is counted six times so there are
\[\frac{60}{6}=10 \, \text{different committees}. \nonumber\]
In both cases we divided the number of permutations by the number of ways to rearrange the people chosen.
The number of permutations of n people taking r at a time is \(P(n,r)\) and the number of ways to rearrange the people chosen is \(r!\). Putting these together we get
\[\begin{aligned}
\frac{n !}{\# \text { ways to arrange r items }} &=\frac{P(n, r)}{r !}=\frac{(n-r) !}{\frac{r !}{1}} \\
&=\frac{n !}{(n-r) !} \cdot \frac{1}{r !} \\
&=\frac{n !}{(n-r) ! r !}
\end{aligned}\]
A combination is a selection of objects in which the order of selection does not matter. The number of combinations of n items taking r at a time is:
\[C(n, r)=\frac{n !}{r !(n-r) !} \label{combination}\]
Note: Many calculators can calculate combinations directly. Look for a function that looks like \(_nC_r\) or \(C(n,r)\) .
A student has a summer reading list of eight books. The student must read five of the books before the end of the summer. In how many ways can the student read five of the eight books?
Solution
The order of the books is not important, only which books are read. This is a combination of eight items taking five at a time.
\[C(8,5)=\frac{8 !}{5 !(8-5) !}=\frac{8 !}{5 ! 3 !}= \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 2 \times 1} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56\]
There are 56 ways to choose five of the books to read.
A child wants to pick three pieces of Halloween candy to take in her school lunch box. If there are 13 pieces of candy to choose from, how many ways can she pick the three pieces?
Solution
This is a combination because it does not matter in what order the candy is chosen.
\[\begin{align*} C(13,3) &=\frac{13 !}{3 !(13-3) !}=\frac{13 !}{3 ! 10 !} \\[4pt] &= \dfrac{13 \times 12 \times 11 \times 10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(10 \times 9\times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)} \\[4pt] &=\frac{13 \times 12 \times 11}{3 \times 2 \times 1} \\[4pt] =\frac{1716}{6}=286 \end{align*}\]
There are 286 ways to choose the three pieces of candy to pack in her lunch.
Note: The difference between a combination and a permutation is whether order matters or not. If the order of the items is important, use a permutation. If the order of the items is not important, use a combination.
Now here are a couple examples where we have to figure out whether it is a permuation or a combination.
A serial number for a particular model of bicycle consists of a letter followed by four digits and ends with two letters. Neither letters nor numbers can be repeated. How many different serial numbers are possible?
Solution
This is a permutation because the order matters.
Use the multiplication principle to solve this. There are 26 letters and 10 digits possible.
\[26 \cdot 10 \cdot \underline{9} \cdot \underline{8} \cdot \underline{7} \cdot \underline{25} \cdot \underline{24}=78,624,000\]
There are 78,624,000 different serial numbers of this form.
A class consists of 15 men and 12 women. In how many ways can two men and two women be chosen to participate in an in-class activity?
Solution
This is a combination since the order in which the people is chosen is not important.
Choose two men:
\[C(15,2)=\frac{15 !}{2 !(15-2) !}=\frac{15 !}{2 ! 13 !}=105 \nonumber\]
Choose two women:
\[C(12,2)=\frac{12 !}{2 !(12-2) !}=\frac{12 !}{2 ! 10 !}=66 \nonumber\]
We want 2 men and 2 women so multiply these results.
\[105(66)=6930 \nonumber\]
There are 6930 ways to choose two men and two women to participate.


8: Comets, Asteroids, and Meteors - The Leftovers of the Solar System
Hundreds of smaller members of the solar system—asteroids and comets—are known to have crossed Earth’s orbit in the past, and many others will do so in centuries ahead. What could we do if we knew a few years in advance that one of these bodies would hit Earth?
To understand the early history of life on Earth, scientists study ancient fossils. To reconstruct the early history of the solar system, we need cosmic fossils—materials that formed when our system was very young. However, reconstructing the early history of the solar system by looking just at the planets is almost as difficult as determining the circumstances of human birth by merely looking at an adult.
Instead, we turn to the surviving remnants of the creation process—ancient but smaller objects in our cosmic neighborhood. Asteroids are rocky or metallic and contain little volatile (easily evaporated) material. Comets are small icy objects that contain frozen water and other volatile materials but with solid grains mixed in. In the deep freeze beyond Neptune, we also have a large reservoir of material unchanged since the formation of the solar system, as well as a number of dwarf planets.
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- 8.1: Asteroids
- The solar system includes many objects that are much smaller than the planets and their larger moons. The rocky ones are generally called asteroids. Ceres is the largest asteroid; about 15 are larger than 250 kilometers and about 100,000 are larger than 1 kilometer. Most are in the asteroid belt between Mars and Jupiter. The presence of asteroid families in the belt indicates that many asteroids are the remnants of ancient collisions and fragmentation.
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- 8.2: Asteroids and Planetary Defense
- Near-Earth asteroids (NEAs), and near-Earth objects (NEOs) in general, are of interest in part because of their potential to hit Earth. They are on unstable orbits, and on timescales of 100 million years, they will either impact one of the terrestrial planets or the Sun, or be ejected. Most of them probably come from the asteroid belt, but some may be dead comets. NASA’s Spaceguard Survey has found 90% of the NEAs larger than 1 kilometer, with none on a collision course with Earth.
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- 8.3: The "Long-Haired" Comets
- Halley first showed that some comets are on closed orbits and return periodically to swing around the Sun. The heart of a comet is its nucleus, a few kilometers in diameter and composed of volatiles and solids. Whipple first suggested this “dirty snowball” model in 1950; it has been confirmed by spacecraft studies of several comets. As the nucleus approaches the Sun, its volatiles evaporate (perhaps in localized jets or explosions) to form the comet’s head or atmosphere.
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- 8.4: The Origin and Fate of Comets and Related Objects
- Oort proposed in 1950 that long-period comets are derived from what we now call the Oort cloud, which surrounds the Sun out to about 50,000 AU (near the limit of the Sun’s gravitational sphere of influence) and contains between \(10^{12}\) and \(10^{13}\) comets. Comets also come from the Kuiper belt, a disk-shaped region beyond the orbit of Neptune, extending to 50 AU from the Sun. Comets are primitive bodies left over from the formation of the outer solar system.
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- 8.5: Meteors
- When a fragment of interplanetary dust strikes Earth’s atmosphere, it burns up to create a meteor. Streams of dust particles traveling through space together produce meteor showers, in which we see meteors diverging from a spot in the sky called the radiant of the shower. Many meteor showers recur each year and are associated with particular comets that have left dust behind as they come close to the Sun and their ices evaporate (or have broken up into smaller pieces).
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- 8.6: Meteorites - Stones from Heaven
- Meteorites are the debris from space that survive to reach the surface of Earth. Meteorites are called finds or falls according to how they are discovered; the most productive source today is the Antarctic ice cap. Meteorites are classified as irons, stony-irons, or stones accordingly to their composition. Most stones are primitive objects, dated to the origin of the solar system. The most primitive are the carbonaceous meteorites that can contain a number of organic (carbon-rich) molecules.
Thumbnail: Comet Hale-Bopp was one of the most attractive and easily visible comets of the twentieth century. It is shown here as it appeared in the sky in March 1997. You can see the comet’s long blue ion tail and the shorter white dust tail. You will learn about these two types of comet tails, and how they form, in this chapter. (credit: modification of work by ESO/E. Slawik).
Contributors and Attributions
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Andrew Fraknoi (Foothill College), David Morrison (NASA Ames Research Center), Sidney C. Wolff (National Optical Astronomy Observatory) with many contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/astronomy ).


2.5: Demand for Inputs
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Up to this point, our emphasis has been on the output decisions of firms. A primary goal has been to establish that firms’ output decisions respond to market price as predicted by the law of supply. However, when firms choose output, they must choose inputs. Input choices are also important to profit maximization. Let us briefly consider the problem of simultaneously choosing an output level and the necessary inputs to achieve it. To do so, it will first be necessary to introduce the idea of a production function. A production function is a mathematical representation of the production technology by which inputs are converted into an output. To keep things simple, consider a production technology with two inputs, \(x_{1}\) and \(x_{2}\) that can be converted into an output, \(q\).
\(q = f(x_{1}, x_{2})\)
Figure \(\PageIndex{1}\) presents a three-dimensional rendering of a production function. Output increases as either of the inputs increases. However, it is common in economics to assume that production functions are concave. This is a mathematical way of saying that production functions exhibit the law of diminishing marginal productivity that was described earlier in the chapter. The example production function in Figure \(\PageIndex{1}\) is concave in that output increases at higher levels of inputs \(x_{1}\) and/or \(x_{2}\) but at a decreasing rate. The production surface shown in Figure \(\PageIndex{1}\) always slopes upward (moving away from the origin) but becomes less steep as increasing amounts of either of the inputs are employed.
Because three dimensions is often hard to work with, a production function is typically rendered in two dimensions using an isoquant map similar to that shown below in Figure \(\PageIndex{2}\). To help you visualize the connection between the three-dimensional rendering in Figure \(\PageIndex{1}\) and the two dimensional rendering in Figure \(\PageIndex{2}\), consider the green plane that intersects the production function in Figure \(\PageIndex{1}\) above. The points where this plane intersects the function represent different combinations of \(x_{1}\) and \(x_{2}\) that could be used to obtain a fixed level of output equal to the elevation of the plane. If you were to look at this intersection directly from above, you would see an isoquant similar to one of the grey curves in Figure \(\PageIndex{2}\) below. In Figure \(\PageIndex{2}\), any point on the same isoquant represents a feasible way to produce a fixed level of output from the two inputs. Different isoquants could be obtained from Figure \(\PageIndex{1}\) by raising or lowering the horizontal plane to correspond to a different levels of output. The word, isoquant, is a combination of the prefix “iso”, meaning equal, and the word “quantity”. Thus, you can think of isoquants as “equal quantity curves”. For example, Points D and E in Figure \(\PageIndex{2}\) would each produce the same output level. Point D uses more of the \(x_{2}\) input and less of the \(x_{1}\) input than does point E. Nevertheless, each point represents a production plan that will achieve the same level of output. Using a conventional geographic orientation to Figure \(\PageIndex{2}\), it can be seen that isoquants in a northeasterly direction indicate higher levels of output.
Figure \(\PageIndex{2}\) also shows isocost lines. Again as the name implies, an isocost line plots out production plans that cost the same. Production cost can be expressed in terms of inputs as follows:
\(c = W_{1}x_{2}\)
where \(W_{1}\) and \(W_{2}\) are the prices of inputs 1 and 2, respectively. Solving this expression algebraically for \(x_{2}\) provides the isocost lines depicted in Figure \(\PageIndex{2}\). These are of the form
\(x_{2} = \dfrac{c}{W_{2}} - \dfrac{W_{1}}{W_{2}}x_{1}\)
Two isocost lines are depicted in Figure \(\PageIndex{2}\). One maps out all plans that provide a cost level of \(c = c*\). To the northeast of this is another isocost line that maps out all plans providing a higher cost level of \(c = \tilde{c})\). In Figure \(\PageIndex{2}\), the production plans represented by points A and D both cost \(c = \tilde{c} )\) because they are each on the isocost line even though more output is produced at D than at A (D lies on an isoquant that is northeast of A).
With these preliminaries, Figure \(\PageIndex{2}\) can be used to visualize the firm’s profit maximization problem. Suppose that at the prevailing output and input prices, the firm’s profit maximizing choice is to set its output at \(q = q*\). This is represented by the blue isoquant in Figure \(\PageIndex{2}\). The firm must choose a production plan that produces \(q*\) at the lowest cost possible. The profit maximizing production plan is represented by point B in Figure \(\PageIndex{2}\). To see why point B is the profit maximizing production plan, consider an alternative production plan like point A. This plan would also result in an output of \(q*\), but the plan itself would not be profit maximizing because the cost at point A is higher (\(\tilde{c}\) as opposed to \(c*\)). At point A, the firm could obtain a lower cost by using less of the \(x_{2}\) input and more of the \(x_{1}\) input. You can know that the production plan at point B meets the cost minimization criterion because it is tangent to the isoquant of the profit maximizing output. At point B, it is not possible to find another production plan that still produces the profit maximizing output level of \(q*\) but at a lower (southwesterly) isocost line.
The fact that the profit maximizing production plan occurs where the isocost line is tangent to the isoquant for the profit maximizing output provides some important economic intuition. Look again at the mathematical expression for the isocost line and notice that it has a slope that is equal to \(-\dfrac{W_{1}}{W_{2}}\). This is simply the negative of the ratio of the market-determined input prices and is the rate at which the economy is willing to trade input \(x_{2}\) for input \(x_{1}\). The slope of the isoquant shows the rate at which the firm can replace \(x_{2}\) with \(x_{1}\) while keeping output constant. Thus, a profit maximizing production plan occurs where the rate of trade-off between the two inputs within the firm is equated to the rate the economy will trade between the two inputs. The slope of the isoquant varies from being steep at low levels of \(x_{1}\) to being quite flat at high levels of \(x_{1}\). The slope of the isoquant is given by \(-\dfrac{MP_{1}}{MP_{2}}\), where \(MP_{1}\) and \(MP_{2}\) are the marginal products of \(x_{1}\) and \(x_{2}\), respectively. At a point like A in Figure \(\PageIndex{2}\), there is a lot of \(x_{2}\) being used but not very much \(x_{1}\). Because of the law of diminishing marginal productivity, the marginal product of \(x_{1}\) will be large relative to the marginal product of \(x_{2}\), thereby causing the slope of the isoquant at point A to be steep (large in absolute value).
We are now ready to discuss the demand for inputs. The demand equation for an input will depend on its own-price, the price of other inputs, and the price of the product being produced. Given the production technology in the figures above, the firm’s demand for \(x_{1}\) could be expressed generally as
\(x_{1} = f(W_{1}, W_{2}, P)\).
In the parlance of Chapter 1, the firm’s demand schedule for the first input would be the relationship between \(x_{1}\) and \(W_{1}\). This schedule would shift if the output price, \(P\), or the price of the other input, \(W_{2}\), were to change.
Demonstration \(\PageIndex{1}\) will be used to help you understand input demand. The top panel of the demonstration presents an isoquant map very similar to that shown in above in Figure \(\PageIndex{2}\), the bottom panel shows this firm’s demand curve for the input \(x_{1}\). As you use this demonstration, note that the profit maximizing production plan always occurs at a point where the isoquant is tangent to the isocost line. This reflects the fact that the firm is simultaneously choosing its inputs to minimize the cost of obtaining the desired output. Step through this demonstration by doing each of the following:
- Focus first on the top panel of the demonstration. Increase only the output price from “baseline” to “high”. You will notice that the profit maximizing quantity shifts in a northeasterly direction to a higher isoquant when you do this. Now, decrease the output price back to the baseline and then to low. You see the isoquant shift in a southwesterly direction to lower levels of output. What you are seeing in the top panel is simply the law of supply. The firm’s profit maximizing production plan involves more output at a higher output price than at a lower output price.
- Set the output price and the price and the price of input 2 to “baseline”. Set the price of input 1 to “high”. Now, gradually drop the price of input 1 through each price level until you reach “low”. As you do, pay attention to the relationship between the top panel and the input demand curve in the bottom panel. The input demand curve in the bottom panel simply reflects the profit maximizing production plans from the top panel. Thus, you see that the demand for the input is derived from the profit maximizing behavior of the firm. Note that this input demand satisfies the law of demand as presented in Chapter 1. At lower input prices more is demanded and vice versa .
- Return all values to their baseline level. Now shift the output price from high to low. What happens to the demand schedule for \(x_{1}\)? You should see it shift. Similarly, increase and decrease the price of input 2. You will similarly see a shift in the demand schedule for \(x_{1}\). The takeaway here is that changes in the output price or the price of other inputs will shift the demand for an input. An increase in output price will always increase the demand for an input, all else equal. In this particular example, an increase (decrease) in the price of input 2, shifts the demand for input 1 inwards (outwards).
- Finally, return all values in the demonstration to their baseline levels. Now set the price of input 1 to “low” and the price of input 2 to “moderately high”. Compare the resulting production plan to the baseline plan, denoted by point B in the top panel of the demonstration. Notice that the new production plan involves a large increase in \(x_{1}\) relative to the baseline. The point to be made here is that the optimal production plan will favor lower-priced inputs. You should have noticed that the slope of the isocost line became flatter, which shifted points of tangency to the right in favor of \(x_{1}\).
Demonstration \(\PageIndex{1}\): Deriving the inverse demand schedule for an input from a firm’s profit maximizing behavior.
Capital and Labor Intensity in Agriculture: A Case Example
Figure \(\PageIndex{3}\) shows two methods of harvesting rice. The photo on the left is from Bhutan the photo on the right is from the United States. The method being used in Bhutan is labor intensive. The method being used in California is capital intensive. Which method is the best?
Photo on the left by Steve Evans from Citizen of the World (Bhutan) CC BY 2.0 , via Wikimedia Commons . Photo on the right by Gary Kramer courtesy of USDA Natural Resources Conservation Service., via Wikimedia Commons .
Given the concepts above, a case can be made that each method is probably best given the prices of the two inputs (labor and capital) in Bhutan and the United States. In Bhutan, labor is inexpensive relative to capital. In the United States, the reverse is true. This can be represented on the isoquant/isocost map in Figure 6. The optimal production plan in Bhutan would occur at a point like B, where the slope of the isoquant is relatively flat to match the small labor to capital price ratio. The isocost line in the United States is much steeper. Consequently, an optimal production plan for US rice harvest would occur at a point where the isoquant is equally steep, such as point A in Figure \(\PageIndex{4}\).