# 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?

• 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! – Robert Lewis Oct 10 '18 at 0:37 ## 3 Answers 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. • So setting the derivative equal to zero and solving I get +/-sqrt(12). Where do I go now? – ovil101 Oct 9 '18 at 22:15 • – qbert Oct 9 '18 at 22:17 • Putting sqrt(12) I get about 83. Does this seem correct? – 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? • 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? – ovil101 Oct 9 '18 at 22:16 • 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. – 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

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

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 $$R= 6$$.