MATH 251 Lecture 16

« previous | Friday, February 24, 2012 | next »

Critical points of $f(x,y)=xy\sin {\left(\pi (x-y)\right)}$ bounded by $y=\pm {\frac {x}{2}}$, $x^{2}-y^{2}=1$, and $x=0$.

${\begin{aligned}{\frac {\partial f}{\partial x}}&=y\sin {\left(\pi (x-y)\right)}+\pi xy\cos {\left(\pi (x-y)\right)}\\{\frac {\partial f}{\partial x}}&=x\sin {\left(\pi (x-y)\right)}-\pi xy\cos {\left(\pi (x-y)\right)}\\{\mbox{both true}}&{\begin{cases}f_{x}=0\rightarrow y\left(\sin {\left(\pi (x-y)\right)}+\pi x\cos {\left(\pi (x-y)\right)}\right)=0\\f_{y}=0\rightarrow x\left(\sin {\left(\pi (x-y)\right)}-\pi y\cos {\left(\pi (x-y)\right)}\right)=0\\\end{cases}}\end{aligned}}$

Therefore either $y=0$ or $\sin {\left(\pi (x-y)\right)}+\pi x\cos {\left(\pi (x-y)\right)}=0$ and $x=0$ or $\sin {\left(\pi (x-y)\right)}-\pi y\cos {\left(\pi (x-y)\right)}$

(0,0) is a critical point.

$x=0,\,y\neq 0$ is not in the domain.

...