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?


Let us parametrize the circumfererence:$x=2\cos(\phi),y=1+2\sin(\phi)$. Then the function
 $$f(\phi):=e^{-2\cos(\phi)}(4\cos^2(\phi)-3)(4\sin^2(\phi)+4\sin(\phi)-2)$$ should be maximized on $[0,2\pi]$.
 The equation $f'(\phi)=0$ is equivalent to
 $$
 32  \sin^3 ( \phi )    \cos ^2(
\phi ) -32\cos ( \phi )   \sin^3
 \left( \phi \right) +
 $$ 
  $$
  32  \cos^2 ( \phi )     \sin ^2( \phi)   +
32 \cos ^3( \phi )  \sin ( \phi
 ) -24 \sin^3 ( \phi )  -
 $$
  $$
  32\,\cos
 \left( \phi \right)   \sin^2 \left( \phi \right)  -16
\,  \cos^2 \left( \phi \right) \sin \left( \phi
 \right) +
 $$ 
 $$
 16\,  \cos ^3\left( \phi \right) -24\,
  \sin^2 \left( \phi \right)  -
  $$
   $$
   8\,\cos \left( \phi
 \right) \sin \left( \phi \right) +12\,\sin \left( \phi \right) -12\,
\cos \left( \phi \right) =0.
$$
Its roots are expressed in terms of polynomials of higher degrees (One of these equals 10.). 
The numerical solution with Maple  produces $\max_{\phi \in [0,2\pi]}f(\phi)=13.75012514$ at $\phi=4.105026133$.
