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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?

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  • $\begingroup$ I edited your post to make the $\LaTeX$ work; remember to surround your math with "\$" signs; thus "\$ \pi \$" is $\pi$ and "\$x^2y\$" is $x^2y$! Cheers! $\endgroup$ – Robert Lewis Oct 10 '18 at 0:37
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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.

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  • $\begingroup$ So setting the derivative equal to zero and solving I get +/-sqrt(12). Where do I go now? $\endgroup$ – ovil101 Oct 9 '18 at 22:15
  • $\begingroup$ see here: jwilson.coe.uga.edu/emt725/Class/Pearman/maxf/max.html $\endgroup$ – qbert Oct 9 '18 at 22:17
  • $\begingroup$ Putting sqrt(12) I get about 83. Does this seem correct? $\endgroup$ – ovil101 Oct 9 '18 at 22:38
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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?

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    $\begingroup$ So substituting y^2=36-x^2 I get 36y-y^3. The derivative is 36-3y^2. Setting it equal i get +/-sqrt(12). Where do I go from here? $\endgroup$ – ovil101 Oct 9 '18 at 22:16
  • $\begingroup$ Good. Now use the second derivative to determine which value of $y$ gives the maximum. (The other is a minimum.) Then you should have your answer. $\endgroup$ – John Oct 10 '18 at 16:37
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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 $R= 6$.

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