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?