<s>[INST] Using Lagrange multipliers to solve for minimum I am having troubles with one part of this homework problem.  Hopefully somebody can help me out:
Find the minimum and maximum values of the function $f(x,y)=x^2+y^2$  subject to the given constraint $x^4+y^4=18$. 
Using Lagrange multipliers, I can easily solve for the maximum:
$f_x(x,y)=2x$ and $f_y(x,y)=2y$.
If we call the second equation $g$, then:
$g_x=4x^3$ and $g_y=4y^3$.
Then we apply the Lagrange multiplier:
$2x=\lambda 4x^3$ and $2y = \lambda 4y^3$
By solving for $x$ and $y$ and plugging in to $g$, we get $\lambda=1/\sqrt{36}=\pm1/6$.
To find the maximum, I will use the positive $1/6$, and solve for $x^2=3$ and $y^2=3$, resulting in a maximum of $6$, which the system spits out as correct.
For a minimum, I originally thought $0$ because $x^2$ and $y^2$ must be positive numbers, but that is not correct.  Then I noticed that since $x^2=1/(2\lambda)$, and $y$ also, then when $\lambda$ is negative that would be the minimum, resulting in $-6$, which is also not correct.
So long story short, how do I find the minimum value in this case?
Thanks!
 [/INST] $$\mathcal{L}(x,y,\lambda)=x^2+y^2-\lambda(x^4+y^4-18)$$
We have
$$\nabla \mathcal{L}(x,y,\lambda) = \left(\begin{array}{c}2x-4\lambda x^3\\2y-4\lambda y^3 \\ x^4+y^4-18 \end{array}\right).$$  Setting this equal to zero gives us $$\frac{1}{2\lambda}=x^2 \;\text{ -or- }\; x=0$$ and $$\frac{1}{2\lambda}=y^2 \;\text{ -or- }\; y=0$$
Clearly both $x$ and $y$ cannot be zero at the same time.  So we have three cases:
Case 1)  $$x^2=y^2=\frac{1}{2\lambda} \;\Longrightarrow\; \frac{1}{4\lambda^2}+\frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \lambda = \pm \frac{1}{6}.$$  Since the function is of $x^2$ and $y^2$ we need only consider one of the two cases; let's pick $\lambda=\frac{1}{6}$ which results in $$f(x,y)=x^2+y^2=\frac{1}{2\lambda}+\frac{1}{2\lambda}=6.$$
Case 2) $$x=0, \qquad y^2=\frac{1}{2\lambda}$$
Putting this into the constraint gives us
$$0+y^4=18\;\Longrightarrow\; \frac{1}{4\lambda^2}=18 \;\Longrightarrow\; \frac{1}{72}=\lambda^2 \;\Longrightarrow\; \lambda = \pm \frac{1}{6\sqrt{2}}.$$
This results in $$f(x,y)=x^2+y^2=0+\frac{1}{2\lambda}=\frac{6\sqrt{2}}{2}=3\sqrt{2}.$$
Case 3) is symmetric with case (2) since the function and constraint is symmetric w.r.t.  $x$ and $y$.
We have enumerated all possible values of $f$, so you can easily see which are maxima and minima.  You will have to use some care if you need to enumerate all minimizers and maximizers, as there are a few cases and some symmetry to consider.
P.S., the accepted capitalization of Joseph-Louis Lagrange's surname is with lower-case `g's.  This is different from some other similar words, e.g., LaGrange County, LaGrange College, etc.  I cannot recommend strongly enough sticking with "Lagrange" for capitalization.
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