Find the global maximum and minimum values

Question

Find the global maximum and minimum values of the function $f(x,y) = e^{−xy}$ subject to the constraint $x^2 + 4y^2 \le 1$.

Answer 1

To find an interior critical point, we would need to find a point so that $$ \frac{\partial}{\partial x}e^{-xy}=-ye^{-xy}=0\qquad\text{and}\qquad\frac{\partial}{\partial y}e^{-xy}=-xe^{-xy}=0 $$ That is, $(x,y)=(0,0)$ is the only interior critical point.

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To find a critical point on the boundary $x^2+4y^2=1$, we would need to find when, for any $\delta x$ and $\delta y$ $$ x\,\delta x+4y\,\delta y=0 $$ implies that $$ -ye^{-xy}\,\delta x-xe^{-xy}\,\delta y=0 $$ This would mean that there is a $\lambda$ so that $$ \left(-ye^{-xy},-xe^{-xy}\right)=\lambda(x,4y) $$ That is, $$ 4y^2=x^2 $$ Combined with $x^2+4y^2=1$, we get the critical points to be $$ (x,y)=\left(\pm\frac1{\sqrt2},\pm\frac1{2\sqrt2}\right) $$ Checking out each of these $5$ critical points shows that the global minimum value of $e^{-1/4}$ occurs at $(x,y)=\left(\frac1{\sqrt2},\frac1{2\sqrt2}\right)$ and $(x,y)=\left(-\frac1{\sqrt2},-\frac1{2\sqrt2}\right)$.