Find the maximum value of
\[f(x,y) = x \sqrt{1 - y^2} + y \sqrt{1 - x^2},\]where $-1 \le x,$ $y \le 1.$
This looks like a constrained optimization problem, where I want to maximize a function of two variables subject to some inequality constraints. I could try to use the method of Lagrange multipliers, but that would involve finding the gradients of $f$ and the constraint function, setting them equal to a scalar multiple, and solving for $x, y,$ and the multiplier. That seems a bit tedious and messy. I wonder if there is a simpler way to approach this problem, maybe by using some geometric or algebraic insight. I notice that the constraint $-1 \le x, y \le 1$ defines a square in the $xy$-plane, and that $f$ is symmetric with respect to both $x$ and $y$. That means I only need to consider one quadrant of the square, say the first quadrant where $0 \le x, y \le 1$, and the maximum value of $f$ there will be the same as in the other three quadrants. I also notice that the function $f$ involves square roots of $1 - x^2$ and $1 - y^2$, which remind me of the Pythagorean theorem and the unit circle. In fact, if I let $x = \cos \theta$ and $y = \sin \theta$, where $0 \le \theta \le \frac{\pi}{2}$, then the constraint is automatically satisfied, and $f$ becomes
\[f(\cos \theta, \sin \theta) = \cos \theta \sqrt{1 - \sin^2 \theta} + \sin \theta \sqrt{1 - \cos^2 \theta} = \cos \theta \cos \theta + \sin \theta \sin \theta = 1.\] This is a constant function, which means it does not depend on $\theta$, and its value is $1$. This means that no matter what point I choose on the boundary of the square, the value of $f$ there will be $1$. Therefore, the maximum value of $f$ is $1$, and it is attained at any point on the boundary of the square. To check this, I can plug in some boundary points, such as $(1, 0), (0, 1), (-1, 0),$ and $(0, -1)$, and verify that they all give $f = 1$. I can also verify that the interior points of the square give smaller values of $f$, by plugging in some random points, such as $(\frac{1}{2}, \frac{1}{2}), (\frac{1}{3}, \frac{2}{3}),$ and $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$, and seeing that they all give $f < 1$. Therefore, I am confident that the maximum value of $f$ is $1$, and I can write down my final answer. 