Find the maximum and minimum of a multivariable function on a circle This question is a continuation from a previous question I recently asked:
Stationary points of a multivariable function
I now have to find the maximum and minimum values of my function on the circle: $x^2 +y^2 = 4$
I'll repeat what I have so you don't have to keep checking back to my previous post:
The function: $f(x,y) = (x^2+2y^2)e^{-x^2-y^2}$
The partial derivatives: $f_x = (-2x(x^2+2y^2)+2x)e^{-x^2-y^2}$, $f_y=(-2y(x^2+2y^2)+4y)e^{-x^2-y^2}$
From here I set up my Lagrange equations (Since I know that that $\nabla f = \lambda \nabla g$ where $g$ is the function of the given circle):
(I also know that $e^{-x^2-y^2}$ simply becomes $e^{-4}$ since $x^2 +y^2 =4$)
$$(-2x(x^2+2y^2)+2x)e^{-4} = 2\lambda x \space (1)$$
$$(-2y(x^2+2y^2)+4y)e^{-4} = 2\lambda y \space (2)$$
$$x^2+y^2 = 4 \space (3)$$
From here I wasn't entirely sure what to do but I tried dividing both sides of $(1)$ by $2x$ to get: $-e^{-4}(x^2+2y^2)+e^{-4} = \lambda$, and similarly divide both sides of $(2)$ by $2y$ to get: $-e^{-4}(x^2+2y^2)+2e^{-4} = \lambda$
So my new Lagrange equations would be:
$$-e^{-4}(x^2+2y^2)+e^{-4} = \lambda \space (4)$$
$$-e^{-4}(x^2+2y^2)+2e^{-4} = \lambda \space (5)$$
$$x^2 +y^2 = 4 \space (3)$$
I'm not even sure if what I did is correct and even so I'm not sure where I would go from here, equating $(4)$ and $(5)$ wouldn't help (At least I don't think it will) and so I'm quite confused on where to go from here or where I've made a mistake.
If anyone can guide me in the right direction or show me a better method it would really help, thanks in advance
 A: Before analyzing your solution, I'll use a simpler method: the function to study is
$$
F(x)=(8-x^2)e^{-4}
$$
for $x\in[-2,2]$. The minimum is $4e^{-4}$ and the maximum is $8e^{-4}$.
If you want to go the hard way, the equations to solve are
$$
\begin{cases}
2x(1-x^2-2y^2)e^{-x^2-y^2}=2\lambda x \\[6px]
2y(2-x^2-2y^2)e^{-x^2-y^2}=2\lambda y \\[6px]
x^2+y^2=4
\end{cases}
$$
Since $x^2+y^2=4$ you can't have both $x=0$ and $y=0$, so there are three cases.
Case 1
$$
\begin{cases}
x=0 \\[6px]
(2-x^2-2y^2)e^{-4}=\lambda \\[6px]
x^2+y^2=4
\end{cases}
$$
This leads to $y=\pm2$.
Case 2
$$
\begin{cases}
(1-x^2-2y^2)e^{-x^2-y^2}=\lambda \\[6px]
y=0 \\[6px]
x^2+y^2=4
\end{cases}
$$
This leads to $x=\pm2$
Case 3
$$
\begin{cases}
(1-x^2-2y^2)e^{-x^2-y^2}=\lambda \\[6px]
(2-x^2-2y^2)e^{-x^2-y^2}=\lambda \\[6px]
x^2+y^2=4
\end{cases}
$$
This has no solution.
Hence you have to consider
$$
f(0,\pm2)=8e^{-4}
\qquad
f(\pm2,0)=4e^{-4}
$$
A: We are given that $x^2+y^2=4$ and we want to minimize $f(x,y)=(x^2+2y^2)e^{-x^2-y^2}$.
Since $x^2+y^2=4$, we know that on this curve,
$$f(x,y)=(4+y^2)e^{-4}$$
This is now just a single variable expression dependent on $y$.
On the curve $x^2+y^2=4$, we have that $-2\leq y\leq 2$. Taking the derivative wrt $y$,
$$\frac{\partial f(x,y)}{\partial y}=2e^{-4}y$$
Hence, there is a critical value when $y=0$
We can then use candidates test and evaluate the function when $y\in\{-2,0,2\}$ to conclude that the minima of $\boxed{4e^{-4}}$ occurs when $y=0$ and the maxima of $\boxed{8e^{-4}}$ occurs when $y=\pm 2$.
