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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?
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For closed form, looks pretty hopeless, because of mixture of polynomial and exponential. – André Nicolas Jun 26 '13 at 12:41
@AndréNicolas Yes, but the second system has no such mixture, "just" polynomials. – Julian Kuelshammer Jun 26 '13 at 12:42
@Amzoti Depends a bit how advanced these numerical techniques are. I would like to present the solution to first year engineer students in the end. So I'd prefer something that one could compute by hand, but could live with, e.g. having a polynomial of high degree where one has to approximate the roots with the help of a computer. – Julian Kuelshammer Jun 26 '13 at 12:54
@JulianKuelshammer: True, maybe it turns out OK. – André Nicolas Jun 26 '13 at 12:57
@Amzoti Thanks, that would of course be an approach, but it should remain an exercise on Lagrange multipliers. – Julian Kuelshammer Jun 26 '13 at 13:00
up vote 1 down vote accepted

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).

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@Amzoti Here is our verdict. – Julian Kuelshammer Jul 26 '13 at 6:53

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.

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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 $\phi=4.105026133$.

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The circumference under consideration can be also parametrized in such way: $$ x=2\frac {2t} {1+t^2}, y= 2 \frac {1-t^2} {1+t^2} +1.$$ Substituting these in the first equation suggested by S.B., you obtain a polynomial equation in $t$. In this way you remain in LMM. – user64494 Jun 26 '13 at 17:51
@ Julian Kuelshammer: I am troubbeled by no feedback of you. I think the civilized people use to give responses to the answers to their questions. – user64494 Jun 27 '13 at 12:36
Sorry, your last comment didn't arrive at me, because of too many spaces. I didn't check your solution, because although it seems nice, it does not meet the requirement to use Lagrange multiplier method, which is what we wanted to practice with our engineer students. Nevertheless thanks for your effort. – Julian Kuelshammer Jul 26 '13 at 6:55

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