Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$

Question

I am trying to use Lagrange multipliers to find the maximum and minimum values of the function $$f(x,y)=e^{-xy}$$ constrained as $$x^2+4y^2=5$$

I began this problem by setting up the Lagrangian:

$$f(x,y) = e^{-xy}$$ $$g(x,y) = x^2+4y^2-5$$ $$L(x,y) = f(x,y) - \lambda g(x,y) = e^{-xy} - \lambda (x^2+4y^2-5)$$

So our equations are:

$$-ye^{-xy} - 2\lambda x = 0$$ $$-xe^{xy} -8\lambda y = 0$$ $$x^2+4y^2-5 = 0$$

Now from the first equation, $e^{-xy} = 2\lambda x/y$. Substituting into the second equation yields: $2x^2 \lambda / y = 8 \lambda y$ or $x^2+4y^2 = 0$. This clearly violates the third equation, meaning this system has no solution. Doe this mean there are no local maxima or minima?

Any guidance is greatly appreciated!

Answer 1

You have some mistakes. Applying Lagrange Multiplier method, you get

$-ye^{-xy} - 2\lambda x = 0$

$-xe^{-xy} -8\lambda y = 0$

$x^2 + 4y^2 - 5 = 0$

From first, $e^{-xy} = - \cfrac{2 \lambda x}{y}$ for $ \ x, y \ne 0$

Substituting in second, $\cfrac{2 \lambda x^2}{y} = 8 \lambda y$,

For $\lambda \ne 0, x^2 = 4y^2 \implies x = \pm 2y$

Plugging into constraint, $x = \pm \sqrt{\dfrac{5}{2}}, \ y = \pm \sqrt{\dfrac{5}{8}}$

Can you take it from here?

Answer 2

Add and subtract twice the first equation to/from the second equation to obtain

$$-(x+2y)(e^{-xy}+4\lambda)=0$$

$$-(x-2y)(e^{-xy}-4\lambda)=0$$

Now you have four critical points corresponding to the intersections of $x=\pm 2y$ with the ellipse.

Answer 3

$\min f$ is at $\max (xy)$

If $g = x y$ is defined on $x^2 + 4 y^2 <= 5$

Then maximum of $g$ occurs at the boundary of the elliptical region, namely the ellipse $x^2 + 4 y^2 = 5$

The semi-major and semi-minor axes are:

$a = \sqrt{5}$

$b = \sqrt{5}/2$

The parameteric equation of the boundary is

$p(t) = ( \sqrt{5} \cos t, \sqrt{5}/2 \sin t )$

$g = x y = 5/4 \sin (2 t)$ , so it is maximum at $t = \dfrac{\pi}{4}, \dfrac{5 \pi}{4}$

and minimum of $g$ occurs at $t = \dfrac{3\pi}{4}, \dfrac{7 \pi}{4}$

Thus $\min f = e^{-5/4} $ and $\max f = e^{5/4}$

The minimum occurs at $( \pm \sqrt{\dfrac{5}{2} } , \pm \sqrt{\dfrac{5}{8}} )$

And the maximum occurs at $( \mp \sqrt{ \dfrac{5}{2} } , \pm \sqrt{ \dfrac{5}{8} } )$

Answer 4

The function $ \ f(x \ , \ y) \ = \ e^{\ -xy} \ $ to be extremized has "diagonal symmetry", which is to say that $ \ f(-x \ , \ -y) \ = \ f(x \ , \ y) \ \ ; $ it also has an anti-symmetric property in that $ \ f(-x \ , \ y) \ = \ f(x \ , \ -y) \ = \ \frac{1}{f(x \ , \ y)} \ \ . $ The elliptical region within which we wish to locate extrema, $ \ x^2 + 4y^2 \ \le \ 5 \ \ , $ has "four-fold" symmetry about the origin, so we expect to find the minima of $ \ f(x \ , \ y) \ $ in the first and third quadrants at the points $ \ ( \pm x \ , \ \pm y ) \ $ and its maxima in the second and fourth quadrants at $ \ ( \pm x \ , \ \mp y ) \ \ . $

While we can perform the usual critical-point analysis in this elliptical region, we may also examine the values of our function on the concentric ellipses $ \ x^2 + 4y^2 \ = \ c^2 \ \ , \ $ with $ \ 0 \ \le \ c^2 \ \le \ 5 \ $ to find the value of $ \ c \ $ where the extrema reach their greatest and smallest values. Due to the symmetries noted, we can just look at each ellipse in the first quadrant, for which $ \ f(x \ , \ y) \ = \ \large{e^{\ -\sqrt{c^2 - 4y^2} \ · \ y }} \ \ . $ This plainly takes on its largest value at the origin, where $ \ c \ = \ 0 \ \Rightarrow \ f(0 \ , \ 0) \ = \ e^{\ 0} \ = \ 1 \ \ . $ We have, for $ \ c \ \neq \ 0 \ \ , $ the derivative $$ \frac{d}{dy} \ [ \ e^{\ -\sqrt{c^2 - 4y^2} \ · \ y} \ ] \ \ = \ \ e^{\ -\sqrt{c^2 - 4y^2} \ · \ y} \ · \ \frac{d}{dy} \ [ \ -\sqrt{c^2 - 4y^2} \ · \ y \ ] $$ and since $ \ f(x \ , \ y) \ \neq \ 0 \ \ $ for real values of the coordinate variables, we find the minimum value of our function from $$ \frac{d}{dy} \ [ \ -\sqrt{c^2 - 4y^2} \ · \ y \ ] \ \ = \ \ -\frac12·\frac{1}{\sqrt{c^2 - 4y^2}}·(-8y) \ · \ y \ + \ -\sqrt{c^2 - 4y^2} \ · \ 1 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ -4y^2 \ \ = \ \ c^2 - 4y^2 \ \ \Rightarrow \ \ y^2 \ \ = \ \ \frac{c^2}{8} \ \ \Rightarrow \ \ x^2 \ \ = \ \ c^2 \ - \ 4·\left(\frac{c^2}{8} \right) \ \ = \ \ \frac{c^2}{2} \ \ . $$

The minima on the concentric ellipses are therefore $ \ f\left(\pm \sqrt{\frac{c^2}{2}} \ , \ \pm \sqrt{\frac{c^2}{8}} \right) \ = \ \large{e^{\ - \sqrt{\frac{c^2}{2}} \ · \ \sqrt{\frac{c^2}{8}} }} \ = \ \normalsize{e^{\ - c^2/4} } \ \le \ 1 \ \ $ and the maxima are $ \ f\left(\pm \sqrt{\frac{c^2}{2}} \ , \ \mp \sqrt{\frac{c^2}{8}} \right) \ = \ \large{e^{\ \sqrt{\frac{c^2}{2}} \ · \ \sqrt{\frac{c^2}{8}} }} \ = \ \normalsize{e^{\ c^2/4} } \ \ge \ 1 \ \ . $

The global extrema are attained by making $ \ c^2 \ $ as large as is permitted on the elliptical region, that is, at its boundary: $$ \mathbf{\text{global minima:}} \ \ \ f\left(\pm \sqrt{\frac{5}{2}} \ , \ \pm \sqrt{\frac{5}{8}} \right) \ \ = \ \ e^{\ - 5/4} \ \ ; $$ $$ \mathbf{\text{global maxima:}} \ \ \ f\left(\pm \sqrt{\frac{5}{2}} \ , \ \mp \sqrt{\frac{5}{8}} \right) \ \ = \ \ e^{\ 5/4} \ \ . $$