<s>[INST] Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariable calculus for this problem. (I've only taken single variable)
The only solution I've seen uses the Cauchy-Schwarz inequality.
 [/INST] Here is an geometric interpretation:  the locus of points $(a,b,c)$ such that $a+b+c = 2$ and $a^2+b^2+c^2 = 12$ is the intersection of a plane and a sphere in $\mathbb R^3$; i.e., it is a circle.  The plane is perpendicular to the line $a = b = c$, and a little visualization leads us to conclude that the extrema of $c$ occur when $a = b$.  In such a case, it becomes straightforward to compute the extrema explicitly: we have  $2a + c = 2$, $2a^2 + c^2 = 12$, hence the difference is $\frac{16}{3}$.
A slightly more rigorous argument involves the coordinate transformation $x = (a+b)/\sqrt{2}$, $y = (a-b)/\sqrt{2}$.  Then our system becomes $c+x\sqrt{2} = 2$, $x^2+y^2+c^2 = 12$, hence $$0 = \frac{3}{2}c^2 - 2c - 10 + y^2.$$  Solving for $c$, we obtain $$c = \frac{2 \pm \sqrt{64-6y^2}}{3},$$ and we observe that these values are greatest/least when $y = 0$.  So the difference is $2\frac{\sqrt{64}}{3} = \frac{16}{3}$.
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