<s>[INST] Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$ 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!
 [/INST] $\min f$ 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 $( \mp \sqrt{ \dfrac{5}{2} } , \pm \sqrt{ \dfrac{5}{8} } )$
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