I meet this question while doing exercises in a textbook. Pre-university math background is assumed.
When $x^2+y^2 \le 4$,what are the maximum and minimum of $f=\sin x \sin y$?
The answer is that maximum is $(\sin(\sqrt 2))^2$ when $x=y=\sqrt 2$, but I don't know how to calculate. My idea:
$f_x=0 \Rightarrow \cos x\sin y=0$
$f_y=0 \Rightarrow \sin x\cos y=0$
Solve the equations, we get $x=\frac{\pi}{2}$, $y=\frac{\pi}{2}$.
But the stationary point $(\frac{\pi}{2},\frac{\pi}{2})$ is not in the domain($x^2+y^2\le 4$), so we will get the maximum/minimum on the boundary of domain. Then I use the lagrange-multiplier, the following equations have to be solved:
First, the Lagrange Function is $L=f+\lambda (x^2+y^2-4)$
Then, we solve the equations
$L_x=0 \Rightarrow \cos x \sin y+\lambda (2x)=0$
$L_y=0 \Rightarrow \sin x\cos y+\lambda(2y)=0$
$x^2+y^2-4=0 $
I take the advantage of the symmetry of the first two equations to get $x=y$ or $x=-y$ (but I don't know if it is right by the symmetry)
In the end, we find $(-\sqrt 2,-\sqrt 2)$ and $(\sqrt 2,\sqrt 2)$ are the maximum points, $(-\sqrt 2,\sqrt 2)$ and $(\sqrt 2,-\sqrt 2)$are the minimum points.
The problem can be solved by elementary tools. While searching for the maximum, we may assume that $\sin x$ and $\sin y$ have the same sign. Since the region is symmetric we may restrict ourselves to $\sin x>0$ and $\sin y>0,$ i.e. $x,y\in [0,2].$ Basing on the inequality $\sqrt{ab}\le {1\over 2}(a+b)$ we get $$\sqrt{\sin x\,\sin y}\le {\sin x+\sin y\over 2}=\sin{x+y\over 2}\,\cos {x-y\over 2}\le \sin{x+y\over 2}$$ We have $x+y\le \sqrt{2}\sqrt{x^2+y^2}\le 2\sqrt{2},$ therefore $${x+y\over 2}\le \sqrt{2}\le {\pi\over 2}$$ As the function $\sin x$ is increasing in the first quadrant, we get $$\sqrt{\sin x\,\sin y}\le \sin\sqrt{2},$$ hence $$\sin x\,\sin y\le (\sin\sqrt{2})^2$$ By analyzing all the inequalites applied on the way, we can notice that equalities hold for $x=y.$ Hence the maximal value is attained at $x=y=\sqrt{2}$ as well as at $x=y=-\sqrt{2}.$
Since the function $\sin x\,\sin y$ is odd with respect to $x$ and with respect to $y$ the minimum is equal $-(\sin\sqrt{2})^2$ and is attained at $x=-y=\sqrt{2}$ and $-x=y=\sqrt{2}.$