What are the extrema and saddle points of f(x, y) = 6 sin x sin y on the interval x,y in[-pi,pi] ?

Nov 5, 2015

$x = \frac{\pi}{2}$ and $y = \pi$

$x = \frac{\pi}{2}$ and $y = - \pi$

$x = - \frac{\pi}{2}$ and $y = \pi$

$x = - \frac{\pi}{2}$ and $y = - \pi$

$x = \pi$ and $y = \frac{\pi}{2}$

$x = \pi$ and $y = - \frac{\pi}{2}$

$x = - \pi$ and $y = \frac{\pi}{2}$

$x = - \pi$ and $y = - \frac{\pi}{2}$

Explanation:

To find the critical points of a $2$-variable function, you need to compute the gradient, which is a vector cointaining the derivatives with respect to each variable:

$\left(\frac{d}{\mathrm{dx}} f \left(x , y\right) , \frac{d}{\mathrm{dy}} f \left(x , y\right)\right)$

So, we have

$\frac{d}{\mathrm{dx}} f \left(x , y\right) = 6 \cos \left(x\right) \sin \left(y\right)$, and similarly

$\frac{d}{\mathrm{dy}} f \left(x , y\right) = 6 \sin \left(x\right) \cos \left(y\right)$.

In order to find the critical points, the gradient must be the zero vector $\left(0 , 0\right)$, which means solving the system

$\left\{\begin{matrix}6 \cos \left(x\right) \sin \left(y\right) = 0 \\ 6 \sin \left(x\right) \cos \left(y\right) = 0\end{matrix}\right.$

which of course we can simplify getting rid of the $6$'s:

$\left\{\begin{matrix}\cos \left(x\right) \sin \left(y\right) = 0 \\ \sin \left(x\right) \cos \left(y\right) = 0\end{matrix}\right.$

This system is solved choosing for $x$ a point which annihilates the cosine, and for $y$ a point which annihilate the sine, and vice versa, so

$x = \pm \frac{\pi}{2}$, and $y = \pm \pi$, and vice versa $x = \setminus \pm \pi$ and $y = \setminus \pm \frac{\pi}{2}$, obtaining $8$ points in total.