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.
 A: 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 $2\frac{\sqrt{64}}{3} = \frac{16}{3}$.
A: here is detail explanation of Will Jagy's solution:
$2(a^2+b^2) \ge (a+b)^2$ when $a=b$ hold "=".
$\iff 24-2c^2\ge 4-4c+c^2\iff (3c-10)(c+2)\ge0 \iff -2\le c \le \dfrac{10}{3}$
A: Both extremes occur when $a=b.$
This is a geometric observation, the plane $a=b$ is a plane of symmetry for both the sphere and the (orthogonal) plane $a+b+c=2.$ Put another way, switching $(a,b)$ does not change anything, you still get a point in the intersection.
So, there is an argument that makes smaller demands on background. The intersection of a plane and sphere is a circle. For any point $(a,b,c),$ whenever $a \neq b,$ there are two points with the same $c$ value, $(a,b,c)$ and $(b,a,c).$ Such points are not along the circle diameter along which the maximum and minimum values of $c$ occur. So, the diameter has $a=b.$ Substitution then gives max $c = 10/3,$ min $c=-2.$ 
Oh, well. By carefully plugging in, one may confirm that this gives your circle as a function of an independent variable $t$
$$  a = \frac{2}{3} -   \frac{4}{\sqrt 3} \cos t -   \frac{4}{ 3} \sin t,    $$
$$  b = \frac{2}{3} +   \frac{4}{\sqrt 3} \cos t -   \frac{4}{ 3} \sin t,    $$
$$  c = \frac{2}{3} +    \frac{8}{ 3} \sin t,    $$
as $a+b+c=2$ and $a^2 + b^2 + c^2 = 12.$ As $\sin t$ varies between $1$ and $-1,$ the largest $c$ gets is $10/3,$ the smallest $-6/3 = -2.$ The extremes occur when $\cos t = 0,$ so $a=b.$
The information used to construct the parametrization above comes from the upper left three rows by three columns of
$$    
 \left(  \begin{array}{rrrrrrrrrr}
  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  2  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  3  &  -1  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  4  &  -1  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  5  &  -1  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  6  &  -1  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  7  &  -1  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  8  &  -1   \\
  1  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  0  &  9   
\end{array}
  \right),
  $$
with columns that are orthogonal, but not orthonormal, eigenvectors for the matrix with all entries equal to $1.$
A: Eliminating $b,$ we have $$a^2+(2-a-c)^2+c^2=12$$
$$\implies 2a^2-2a(2-c)+2c^2-4c-8=0\iff a^2-a(2-c)+c^2-2c-4=0$$ which is a Quadratic Equation in $a$
As $a$ is real, the discriminant must be $\ge0$ 
$$\implies (2-c)^2-4(c^2-2c-4)\ge0$$
$$\implies -3c^2+4c+20\ge0\iff3c^2-4c-20\le0$$
Now, we know if $(x-\alpha)(x-\beta)\le0$ with $\alpha\le\beta$
we can write $\alpha\le x\le\beta$
If $\alpha,\beta(>\alpha)$ are the roots of $3c^2-4c-20=0,$ we need $\displaystyle\beta-\alpha=+\sqrt{\left(\beta+\alpha\right)^2-4\beta\alpha}$
we have $\displaystyle\beta+\alpha=\frac43$ and $\displaystyle\beta\alpha=-\frac{20}3$
