Maximum value on a circle I need to find the maximum value of a function on a circle: Let $C$ denote the circle of radius $6$ centered at the origin in the $xy$-plane. Find the maximum value of $x^2y$ on $C$. Where do I even start with this?
 A: Hint: For $(x,y)$ on the circle of radius $6$, we have 
$$
x^2=36-y^2
$$
So you can find a single variable function to maximize.
A: If you have a circle of radius $6$ centered at the origin, then you know that every point on the circle satisfies $x^2+y^2=6^2$.
This lets you express your function in terms of one variable, which you can then take derivatives of to find the maxima.
Can you try from here?
A: What about polars?  If we allow the circle $C(R)$ to be of arbitrary positive radius $R$, then for $(x, y) \in C(R)$,
$x = R\cos \theta, \tag 1$
$y = R \sin \theta, \tag 2$
so that
$x^2y = R^3 \cos^2 \theta \sin \theta; \tag 3$
this will take on a maximum value whenever $\cos^2 \theta \sin \theta$ does; we have
$\dfrac{d(\cos^2 \theta \sin \theta)}{d\theta} = -2\cos \theta \sin^2 \theta + \cos^3 \theta; \tag 4$
we set this to $0$ and obtain
$2\cos \theta \sin^2 \theta = \cos^3 \theta; \tag 5$
now,
$\cos \theta = 0 \Longrightarrow \theta = \pm \dfrac{\pi}{2}, \tag 6$
and also
$\cos^2 \theta \sin \theta = 0; \tag 7$
next we bear these facts in mind whilst we investigate
$\cos \theta \ne 0 \Longrightarrow 2\sin^2 \theta = \cos^2 \theta \Longrightarrow \tan^2 \theta = \dfrac{1}{2} \Longrightarrow \tan \theta = \pm \dfrac{1}{\sqrt 2}; \tag 8$
it is easy to see that the point in the first quadrant on the unit circle with $\tan \theta = 1/\sqrt 2$ is 
$(\cos \theta, \sin \theta) = (x, y) =\left ( \dfrac{\sqrt 2}{\sqrt 3}, \dfrac{1}{\sqrt 3} \right ); \tag 9$
then it is also easy to see that the three other points on $C(1)$ such that $\tan \theta = \pm 1 / \sqrt 2$ are
$(\cos \theta, \sin \theta) = (x, y) = \left ( -\dfrac{\sqrt 2}{\sqrt 3}, \dfrac{1}{\sqrt 3} \right ), \;  \left ( \dfrac{\sqrt 2}{\sqrt 3}, -\dfrac{1}{\sqrt 3} \right ), \;  \left ( -\dfrac{\sqrt 2}{\sqrt 3}, -\dfrac{1}{\sqrt 3} \right ); \tag{10}$
to these four critical points $(\pm \sqrt 2 / \sqrt 3, \pm 1 \sqrt 3)$ of $\cos^2 \theta \sin \theta$ on $C(1)$, we must add the two found above where $\theta = \pi / 2, - \pi / 2 \equiv 3 \pi / 2$ and our function $\cos^2 \theta \sin \theta = 0$; since $\cos^2 \theta \ge 0$ everywhere, we look to the factor $\sin \theta$ as potentially determinative of the types of the critical points we have found--whether maxima or minima.  For example, the points $(\pm \sqrt 2 / \sqrt 3, 1 / \sqrt 3)$ are both clearly local maxima since there and there only is $x^2y$ is positive; similarly the points $(\pm \sqrt 2 / \sqrt 3, -1 / \sqrt 3)$ are local minima since there $x^2y = \cos^2 \theta \sin \theta < 0$; finally, the point where $\theta = \pi / 2$, $\cos \theta = \cos^2 \theta \sin \theta = 0$ is seen to be a local minimum, since $\cos^2 \theta \sin \theta > 0$ slightly to either side of it; that is, for $0 < \vert \theta  - \pi / 2 \vert < \epsilon$ for $\epsilon > 0$ sufficiently small; and a similar argument shows $\theta = -\pi / 2 \equiv 3\pi / 2$ is a local maximum.
Thus we have found all the critical points of $x^2y = \cos^2 \theta \sin \theta$ on $C(1)$, of these three are maxima as given above.  
On the circle $C(R)$ the function $x^2y$ becomes $R^3 \cos^2 \theta \sin \theta$; the values of $x$ and $y$ scale with $R$; maxima and minima occur for the same $\theta$ values, however.
I leave it to the reader to supply the numerical details for the case $R= 6$.
