MATH 251 Lecture 16

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

$\begin{array}{rl}\frac{\partial f}{\partial x}& =y\sin (\pi (x-y))+\pi xy\cos (\pi (x-y))\\ \frac{\partial f}{\partial x}& =x\sin (\pi (x-y))-\pi xy\cos (\pi (x-y))\\ \text{both true}& \{\begin{array}{l}{f}_x=0\to y(\sin (\pi (x-y))+\pi x\cos (\pi (x-y)))=0\\ f_y=0\to x(\sin (\pi (x-y))-\pi y\cos (\pi (x-y)))=0\\ \end{array}\end{array}$

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

(0,0) is a critical point.

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

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