Determine all the extrema of a function subject to a non-linear constraint. QUESTION
Determine all extrema of the function
$$f(x,y) = x+ 2y $$ subject to $$x^2 + y^2 - 80 = 0$$
ATTEMPT
I don't think I understand what I'm supposed to do. This was in a test and I ended up trying to "graphically"or ïntuitively" find out how the $f(x,y)$ would behave in a circle of radius $\sqrt80$
Which left me with some pretty random numbers that turned out to also be wrong. 
What could I have done differently? (If possible could I get a bit of a detailed explanation or some links to that info.)
Lagrange Multipliers: I tried that too on the paper and it also went horribly wrong (but I did get some points, though) but I couldn't quite figure it out.
 A: This looks like a usual problem for Lagrange multipliers. You define a Lagrange function via
$$L(x,y,\lambda)= x+2y + \lambda \cdot (x^2+y^2-80)$$
Now you look for critical points of the function $L$.
But here is the lagrange part more explicit. A critical point is a point where the gradient of $L$ is 
zero, as 
\begin{align*}
\frac{\partial L}{\partial x} &= 1+ 2x\lambda \\
\frac{\partial L}{\partial y} &= 2+ 2y\lambda\\
\frac{\partial L}{\partial \lambda} &= x^2+y^2-80 
\end{align*}
we have to solve 
\begin{align*}
0&= 1+2x\lambda \\
0&= 2+2y \lambda \\
0&= x^2+y^2 -80
\end{align*}
Taking the first line times 2 we see that 
$$ 4 x \lambda = 2y \lambda, $$ 
and as $\lambda$ surely isn't $0$ we have 
$$y=2x.$$
Using this information in the third equation we have 
$$ 80 = x^2+ 4x^2 \iff 16=x^2$$
so you have to check $x=\pm 4$. 
Via compactness you know one is a minimum and one is a maximum. So the one where $x$ (and hence $y$ ) is positive must be the maximum.
A: Your initial geometric attempt was good too.
The objective function has a gradient equal to (1,2), so the extrema will lie at the intersection between the circle of radius $\sqrt{80}$ and the line $y-2x=0$. 
Substituting for $y$($=2x$) in $x^2+y^2=80$ yields $x=\pm4$.
