Global extrema of $f(x,y)= e^{-4x^2-9y^2}(2x+3y)$ on the ellipse $4x^2+9y^2 \leq 72$

Question

I'm asked to find the global extremas of the following function :$$f(x,y)= e^{-4x^2-9y^2}(2x+3y)$$ on the ellipse $4x^2+9y^2 \leq 72$

So for the inside, I have $\nabla f=(2e^{-4x^2-9y^2}+(-16x^2-24xy)e^{-4x^2-9y^2}, 3e^{-4x^2-9y^2}+(-36xy-54y^2)e^{-4x^2-9y^2})= (0, 0)$. I get two equations $$-16x^2-24xy+2=0$$ $$-54y^2-36xy+3=0$$ Adding $\frac{-2}{3}$ of the second to the first gives $-16x^3+36y^2=0$,$\to$ $9y^2=4x^2$ $\to$ $ y = \pm \frac{2x}{3}$. Putting that in our original function, I get $e^{-8x^2}(4x)$ and $e^{-8x^2}(0)=0$
So deriving the first (because the second gives $0$) and setting it equal to $0$, I get $x=\pm \frac{1}{16}$ implies $y=\pm \frac{1}{24}$

For the border, I use Lagrange : $$\nabla f = \lambda \nabla g$$ and get : $$-2 e^{-4 x^2 - 9 y^2} (8 x^2 + 12 x y - 1) = \lambda 8x $$ and $$e^{-4 x^2 - 9 y^2} (-36 x y - 54 y^2 + 3)=\lambda 18y$$ So $$\frac{-2(8x^2+12xy-1)}{8x}= \frac{(-36xy-54y^2+3)}{18y}$$ After rearranging, I get $-24x^2+12xy=-54y^2+1$ which seems a bit ugly. According to WolframAlpha, the solution for that is $y=\frac{2x}{3}$ which is one of the value (the other being $-\frac{2x}{3}) $ for the inside of the region that I got before, which I find a bit strange. So my questions are :

1) Is what I did until now even correct ?
2) If not, what did I do wrong and how should I proceed instead ?
3) If yes, how should I proceed now ?
4) Also for the inside values I got (i.e. $x=\pm \frac{1}{16}$, $y =\pm \frac{1}{24} $) : I just need to plug them into the function and at the end find the greatest and smallest values from all candidates, is that correct (I'm quite sure, just want to get a confirmation) ?

Thanks for your help !

Answer 1

From the lagrangian

$$ L(x,y,\mu,\epsilon) = f(x,y) + \mu(g(x,y)-72+\epsilon^2) $$

here $\epsilon$ is a slack variable to transform the inequality constraint into an equality.

The stationary conditions are

$$ \left\{ \begin{array}{rcl} 8 \lambda x-8 (2 x+3 y) x+2& = & 0 \\ 18 \lambda y-18 (2 x+3 y) y+3& =& 0 \\ \epsilon ^2+4 x^2+9 y^2-72 & = & 0 \\ 2 \epsilon \lambda & = & 0 \\ \end{array} \right. $$

here $\lambda = \mu e^{4x^2+9y^2}$. After solving we obtain

$$ \begin{array}{ccccc} x & y & \lambda & \epsilon & f(x,y) \\ -3 & -2 & -\frac{143}{12} & 0 & -\frac{12}{e^{72}} \\ -\frac{1}{4} & -\frac{1}{6} & 0 & \sqrt{\frac{143}{2}} & -\frac{1}{\sqrt{e}} \\ \frac{1}{4} & \frac{1}{6} & 0 & \sqrt{\frac{143}{2}} & \frac{1}{\sqrt{e}} \\ 3 & 2 & \frac{143}{12} & 0 & \frac{12}{e^{72}} \\ \end{array} $$

NOTE

Solutions with $\epsilon = 0$ are solutions at the boundary and solutions with $\epsilon \ne 0$ are interior at the feasible region.

Attached two plots. The first shows the location for the solution points and the second shows a detail for the two internal solutions.

Answer 2

Okay, with the hints, I will try to answer my own question, maybe this will help someone in the future looking at this question.

If we substitute $\hat{x}=2x$ and $\hat{y}=3y$, we get $e^{-\hat{x}-\hat{y}}(\hat{x}+\hat{y})$ Now, if we take the partial derivatives of that, set them equal to 0 and solve, we get finally $\hat{y}=\pm \hat{x}$. Now, if we subtitute that into our original function and take the derivative, we get $\hat{x}=\frac{1}{2}$, so $x=\frac{1}{4}$
$\hat{y}=\pm \hat{x}$ can be rewritten as $y=\pm \frac{2}{3}x$. So now we get $y=\frac{1}{6}$ and $y=\frac{-1}{6}$ as in Cesareo's answer.

On the rand, if we parametrize using polar coordinates with $r=\sqrt{72}=6\sqrt{2}$,
$\hat{x}=rcos(\theta), \hat{y}=rsin(\theta)$, plug that into our function, take the derivative (now in polar coordinates) with respect to $\theta$ (remember that $r$ is obviously fixed) and set it equal to zero, we get $\theta = \frac{\pi}{4}$ and $\theta = \frac{5 \pi}{4}$. So $\hat{x}= r cos(\theta)$ and $\hat{y}=rsin(\theta)$. So we get $\hat{x}=\pm6=\hat{y}$. So we get $x=\frac{\hat{x}}{2}=\pm3$ and $y=\frac{\hat{y}}{3}=\pm2$, again as in Cesareo's answer.

Now, we would just have to plug our different values into our function to see where the global maximum/ minimum is.

Maybe this will help someone in the future.