Lagrange multiplier method, find maximum of $e^{-x}\cdot (x^2-3)\cdot (y^2-3)$ on a circle

Question

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:

1. Do you have any ideas how to solve either of the systems?
2. 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?

Answer 1

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).

Answer 2

You can eliminate $e^{-x}$ and $\lambda$ from the first two equations. Then you have $$(y-1)(y^2-3)(-x^2+2x+3)=2x(x^2-3)y\\x^2+(y-1)^2=4.$$ I'm not sure if you can get explicit solutions, but multiplying the first equation by $y-1$ and then using the second equation may help to simplify the equations.

Answer 3

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 $\phi=4.105026133$.

Answer 4

The system of Lagrange equations can be re-arranged into $$ e^{-x}·(y^2 - 3)·(x^2-2x-3) \ = \ \lambda· 2x \ \ \ , \ \ \ e^{-x}·(x^2-3)·2y \ = \ \lambda · 2(1-y) $$ $$ \Rightarrow \ \ \lambda \ \ = \ \ \frac{(x + 1)·(x - 3)·(y^2 - 3)}{2x} \ \ = \ \ \frac{(x^2 - 3)·y}{1 - y} \ \ ; \ $$ "cross-multiplication" (with $ \ x \ \neq \ 0 \ \ , \ \ y \ \neq \ 1 \ ) \ $ leaves us with a rather unpromising fifth-degree bivariate polynomial. It is not so much that $ \ f(x,y) \ $ is an exponential factor times a polynomial that complicates matters, as it is that the function only has symmetry about the $ \ x-$axis, while the constraint circle has its symmetry about the $ \ y-$axis. There is little in the situation that permits simplification. Below is a diagram showing the circle against the regions where $ \ f(x,y) \ $ is positive [green] or negative [red]; the function is zero on the lines $ \ x \ = \ \pm \sqrt3 \ \ , \ \ y \ = \ \pm \sqrt3 \ \ . $

Because portions of the level-curves $ \ e^{-x}·(x^2 - 3)·(y^2 - 3) \ = \ c \ $ can appear in these regions with only symmetry about the $ \ x-$axis, there are multiple tangencies possible with the constraint circle:

$$ \begin{array}{ccc} \mathbf{\text{x}} & \mathbf{\text{y}} & \mathbf{\text{f(x,y)} } \\ -0.72 & +2.89 & -26.6 \\ -1.98 & +0.73 & -16.6 \\ +1.98 & +0.69 & -0.32 \\ +1.80 & +1.87 & +0.02 \\ -1.80 & +1.87 & +0.73 \\ -1.14 & -0.64 & +13.75 \\ \end{array} $$

The "loop" of the level-curves "to the right" of the constraint circle emerges at small negative values of $ \ c \ $ is "centered" on $ \ x \ = \ 3 \ \ ; \ $ the "loop" within the circle for positive values of $ \ c \ $ contracts as $ \ c \ $ increases, ultimately shrinking to the point $ \ (-1 \ , \ 0 ) \ $ and vanishing for $ \ c \ > \ \approx +16.30 \ \ . $

One is pretty much faced with obtaining the information about these features computationally. The last graph confirms the result for the maximum value $ \ 13.75 \ $ for $ \ f(x,y) \ $ on the constraint circle found by user64494 .

Centering the circle on the origin instead doesn't do much to reduce the difficulty of solving the Lagrange equations. Something that does become simpler is that having symmetry about the origin for the constraint curve now gives us pairs of tangency points $ \ (x \ , \ \pm y) \ $ or tangent points on the $ \ x-$axis. The number of solutions is reduced, but must still largely be found computationally.

For the constraint $ \ x^2 + y^2 \ = \ 4 \ \ , \ $ we obtain

$$ \begin{array}{ccc} \mathbf{\text{x}} & \mathbf{\text{y}} & \mathbf{\text{f(x,y)} } \\ -2 & 0 & -3e^2 \approx -22.17 \\ -0.34 & \pm 1.97 & -3.57 \\ +2 & 0 & 3e^{-2} \approx -0.41 \\ -1.47 & \pm 1.35 & +4.30 \\ \end{array} $$

For the choice you ultimately made, using a constraint circle $ \ x^2 + y^2 \ = \ 1 \ \ $ places it entirely within the central positive region for $ \ f(x,y) \ \ . \ $ Here, because the "loop" of the level-curves is "centered" on the $ \ x-$ axis, the only possibilities for points of tangency are also on that coordinate axis, leaving us with just two solutions:

$$ \begin{array}{ccc} \mathbf{\text{x}} & \mathbf{\text{y}} & \mathbf{\text{f(x,y)} } \\ +1 & 0 & 6e^{-1} \approx +2.21 \\ -1 & 0 & 6e \approx +16.31 \\ \end{array} $$

The tangent point at $ \ (-1 \ , \ 0 \ ) \ $ is the location where that portion of the level curve "shrinks to a point", and thus represents the maximum value of $ \ f(x) \ $ in the central positive region.