Hello pheobee,
We are given the function:
$$f(x,y)=3\sin(x)\sin(y)$$ where $-\pi<x,y<\pi$.
First, to find the critical points, we must solve the simultaneous system:
$$f_x(x,y)=3\cos(x)\sin(y)=0$$
$$f_y(x,y)=3\sin(x)\cos(y)=0$$
This implies by addition:
$$\sin(x)\cos(y)+\cos(x)\sin(y)=\sin(x+y)=0$$
And by subtraction:
$$\sin(x)\cos(y)-\cos(x)\sin(y)=\sin(x-y)=0$$
Hence, we may state:
$$x\pm y=k\pi$$
Now, given the first partials, and the stated domains, we obtain from this the following critical points:
$$(x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right),\,\left(-\frac{\pi}{2},\frac{\pi}{2} \right),\,(0,0),\,\left(\frac{\pi}{2},-\frac{\pi}{2} \right),\,\left(\frac{\pi}{2},\frac{\pi}{2} \right)$$
Next, we may use the second partials test to determine the nature of the potential extrema associated with these 5 critical points.
First, we compute:
$$f_{xx}(x,y)=-3\sin(x)\sin(y)$$
$$f_{yy}(x,y)=-3\sin(x)\sin(y)$$
$$f_{xy}(x,y)=3\cos(x)\cos(y)$$
And we define:
$$D(x,y)\equiv f_{xx}(x,y)f_{yy}(x,y)-\left(f_{xy}(x,y) \right)^2$$
Using our second partials, we then find:
$$D(x,y)=9\left(\sin^2(x)\sin^2(y)-\cos^2(x)\cos^2(y) \right)$$
So now, we analyze the critical points:
1.) $$(x,y)=\left(-\frac{\pi}{2},-\frac{\pi}{2} \right)$$
$$D(x,y)=9>0$$
$$f_{xx}(x,y)=-3<0$$
We conclude this point is at a relative maximum.
The value of the function at this point is:
$$f(x,y)=3$$
2.) $$(x,y)=\left(-\frac{\pi}{2},\frac{\pi}{2} \right)$$
$$D(x,y)=9>0$$
$$f_{xx}(x,y)=3>0$$
We conclude this point is at a relative minimum.
The value of the function at this point is:
$$f(x,y)=-3$$
3.) $$(x,y)=(0,0)$$
$$D(x,y)=-9<0$$
We conclude this point is not at an extremum (saddle point).
The value of the function at this point is:
$$f(x,y)=0$$
4.) $$(x,y)=\left(\frac{\pi}{2},-\frac{\pi}{2} \right)$$
$$D(x,y)=9>0$$
$$f_{xx}(x,y)=3>0$$
We conclude this point is at a relative minimum.
The value of the function at this point is:
$$f(x,y)=-3$$
5.) $$(x,y)=\left(\frac{\pi}{2},\frac{\pi}{2} \right)$$
$$D(x,y)=9>0$$
$$f_{xx}(x,y)=-3<0$$
We conclude this point is at a relative maximum.
The value of the function at this point is:
$$f(x,y)=3$$
And so we may conclude:
$$f_{\min}=-3$$
$$f_{\max}=3$$
Here is a plot of the given function on the stated domain, showing all 5 points:
View attachment 2018
Here is a link to the program I used to plot the function:
z=3sin(x)sin(y) where x=-pi to pi,y=-pi to pi - Wolfram|Alpha
