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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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?

  • 1
    $\begingroup$ 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

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

$2\cos \theta \sin^2 \theta = \cos^3 \theta; \tag 5$


$\cos \theta = 0 \Longrightarrow \theta = \pm \dfrac{\pi}{2}, \tag 6$

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 $R= 6$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.