Lagrange multiplier method, find maximum of $e^{-x}\cdot (x^2-3)\cdot (y^2-3)$ on a circle I attempted to design an exercise for my engineer students and couldn't solve it myself. Maybe here are some experts in calculus who have some better tricks than I do:

The exercise would be to find the maxima of $e^{-x}(x^2-3)(y^2-3)$ on the circle $x^2+(y-1)^2=4$.

Now using the Lagrange multiplier method this amounts to solving the following system of equations:
$$\begin{align*}
e^{-x}(y^2-3)(-x^2+2x+3)+\lambda\cdot 2x&=0\\
e^{-x}(x^2-3)(2y)+\lambda\cdot 2(y-1)&=0\\
x^2+(y-1)^2&=4
\end{align*}$$
I did not succeed to find the solutions and also my standard online calculor didn't.
Now I thought, this is partly because of the $e^{-x}$-term, so it would be good if one could eliminate it. Noting that $x^2+(y-1)^2-4=0$ iff $e^{-x}\cdot (x^2+(y-1)^2-4)=0$ we can instead use Lagrange multipliers on this condition. This amounts to solving the easier system:
$$\begin{align*}
(y^2-3)(-x^2+2x+3)+\lambda(-x^2+2x-(y-1)^2+4)&=0\\
(x^2-3)y+\lambda(y-1)&=0\\
x^2+(y-1)^2-4&=0
\end{align*}$$
In fact I could still not solve it, but the computer could (but the form is not very nice).
So the question is now two-fold:


*

*Do you have any ideas how to solve either of the systems?

*If not, do you have any ideas how to tweak it a little bit (preferrable on the circle condition side) so that it becomes easier to solve?


Here is what we did:
The problem with the stated optimization is that it has too many critical points (for example one sees that when looking at the set of zeros of that function).
So our answer to question 2 was as follows: We moved the circle to maximize on, so that it does not meet the set of zeros of the function we want to maximize, i.e. we took the unit circle $x^2+y^2=1$. Then, when applying the Lagrange multiplier method, one arrives at the two "obvious" solutions $x=\pm 1, y=0$ or the critical points must be solutions of a polynomial equation in $x$ of degree $4$. But for this polynomial equation one can prove that it has no real roots (e.g. by using the formula for polynomial equations of degree $4$, using a computer, or (and that is what we presented) by bounding the values of that polynomial equation (under the restriction $|x|\leq 1$) away from zero).
