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.
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.