global extrema of $(x,y)\mapsto xye^{-x^2-y^2}$ on $\{(x,y)\in\ \mathbb{R}^2\,:\,x^2+y^2=\frac{1}{2}\}$

Question

Find the global extrema of $(x,y)\mapsto xye^{-x^2-y^2}$ on $\{(x,y)\in\ \mathbb{R}^2\,:\,x^2+y^2=\frac{1}{2}\}$ using Lagrangian multipliers.

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My approach: Set $L(x,y,\lambda):=xye^{-x^2-y^2}-\lambda(x^2+y^2-\frac{1}{2})$. It follows that $DL(x,y,\lambda)=\begin{pmatrix}e^{-x^2-y^2}\cdot y\cdot (1-2x^2)-2\lambda x \\ e^{-x^2-y^2}\cdot x\cdot (1-2y^2)-2\lambda y \\ -(x^2+y^2-\frac{1}{2}) \end{pmatrix}^T$.

Question: How do I find the $(x,y,\lambda)$ with $DL(x,y,\lambda)=(0,0,0)$?

Answer 1

Your system is$$\left\{\begin{array}{l}e^{-x^2-y^2}y(1-2x^2)=2\lambda x\\e^{-x^2-y^2}x(1-2y^2)=2\lambda y\\x^2+y^2=\frac12.\end{array}\right.$$If $(x,y,\lambda)$ is a solution of this system, then, since $x^2+y^2=\frac12$, which implies that $1-2x^2=2y^2$ and that $1-2y^2=2x^2$, the system becomes$$\left\{\begin{array}{l}\frac2{\sqrt e}y^3=2\lambda x\\\frac2{\sqrt e}x^3=2\lambda y\\x^2+y^2=\frac12.\end{array}\right.$$Clearly, there is no solution of the system with $\lambda=0$. If $\lambda>0$, the only solutions of the system which consists of the first two equations are$$(x,y)=\pm\left(\sqrt[4]e\sqrt\lambda,\sqrt[4]e\sqrt\lambda\right)$$and, if $\lambda<0$, the only solutions of the system which consists of the first two equations are$$(x,y)=\pm\left(\sqrt[4]e\sqrt{-\lambda},-\sqrt[4]e\sqrt{-\lambda}\right).$$Now, using the third equation, you get that the solutions of your system are$$\left(-\frac{1}{2},\frac{1}{2},-\frac{1}{4\sqrt{e}}\right),\ \left(\frac{1}{2},-\frac{1}{2},-\frac{1}{4\sqrt{e}}\right),\ \left(-\frac{1}{2},-\frac{1}{2},\frac{1}{4\sqrt{e}}\right)\text{, and }\left(\frac{1}{2},\frac{1}{2},\frac{1}{4\sqrt{e}}\right).$$

Answer 2

This can be solved easily by inspection and AM-GM inequality: $xy \leq \frac {x^{2}+y^{2}} 2$ so an upper bound is $\frac 1 4 e^{-\frac 1 2}$. This value is attained when $x=y=\frac 1 2$. Can you now show that the global minimum is $-\frac 1 4 e^{-\frac 1 2}$?