Calculus Made Easy Exercises IX Question 8(a): maximize volume of cylinder inscribed in a cone

Question

Inscribe in a given cone, the height of which is equal to the radius of the base, a cylinder whose volume is a maximum.

I'm stuck. The answer key says the cylinder's height should be $\frac23$ the radius of the base of the cone, but the answer I'm getting is $\frac13$.

The volume of a cylinder is $\pi r^2h$, where $h$ is the height and $r$ is the radius of its base. Since the inscribing cone in this example has height equal to the radius of its own base, we know by similar triangles that any unit of height "added" to the cylinder is "taken" from the the radius of its base. Therefore, the volume $V$ of the inscribed cylinder is $$ \pi h(r-h)^2,$$ where $r$ is the radius of the cone and $h$ is the height of the inscribed cylinder. By the product rule,

$$\frac{dV}{dh} = \pi(r-h)^2 - 2\pi h(r-h).$$

Setting $\frac{dV}{dh}$ equal to 0, we get

$$0 = \pi(r-h)^2 - 2\pi h(r-h)$$ $$2\pi h(r-h) = \pi(r-h)^2$$ $$2h = r - h$$ $$h = \frac13r.$$

Please help!

Answer 1

Let the cone's radius be $r$, and suppose the radius of the cylinder was $x$. Then the height of the cylinder can be determined to be $r-x$ using similar triangles in a triangular cross-section of the cone through the apex. Then the volume of the cylinder is

$$\begin{aligned} V &= \pi x^2(r-x)\\ \dfrac{\mathrm{d}V}{\mathrm{d}x} &= 2\pi rx - 3\pi x^2\\ 0 &= 2\pi rx - 3\pi x^2\\ \pi x(3x - 2r) &= 0\\ x &= 0, \frac{2}{3} r \end{aligned}$$

Trivially $x=0$ doesn't satisfy. Thus, the volume of the cylinder is maximised when the height is $\frac{1}{3}r$.

This agrees with your answer, so I suspect the answers are incorrect. We can indeed verify by checking the volume when $x=\frac{1}{3}r$ (your answer key's value for the cylinder's radius), and when $x=\frac{2}{3}r$ (our calculated value for the maximising radius).

$$x=\frac{1}{3}r \implies V = \frac{2\pi}{27}r^3$$

$$x=\frac{2}{3}r \implies V = \frac{4\pi}{27}r^3$$

Clearly, our value gives the larger volume, so we can indeed confirm the answer key is incorrect.

Answer 2

With the help of Sharky Kesa's answer I figured out the issue. I solved for the cylinder's hight and misunderstood the answer key, which gives its radius. The cylinder of maximum volume has radius and height that are $\frac23$ and $\frac13$ of the cone's radius, respectively.

Thanks for the help.