A cylinder inscribed in a cone · source reference

Printed question

A circular cylinder is inscribed in a right circular cone of base radius $2c$ and vertical height $c$. One of the circular ends of the cylinder lies on the base of the cone and the other end is in contact with the inner surface of the cone. If the cylinder has base radius $x$, find expressions for the volume $V$ and the total surface area $S$ of the cylinder in terms of $c$ and $x$. If $c$ is fixed and $x$ varies, find the maximum value of $V$. [8]

Prove that $S$ cannot exceed $8\pi c^2$. [2]

[2010/NYJC/Prelim/II/Q2]

Printed working

Let $h$ be the cylinder height. Similar triangles give $c/(2c)=(c-h)/x\Rightarrow h=c-x/2$.

$$V=\pi x^2(c-x/2)=\frac{\pi x^2}2(2c-x).$$ $$S=2\pi x^2+2\pi x(c-x/2)=2\pi x^2+2\pi xc-\pi x^2=\pi x(x+2c).$$ $$\frac{dV}{dx}=\frac{\pi x^2}2(-1)+\pi x(2c-x)=\pi x(2c-x-x/2)=\frac{\pi x}2(4c-3x).$$

At maximum/minimum, $dV/dx=0$, so $x=4c/3$ or $x=0$ (rejected).

$$\frac{d^2V}{dx^2}=\frac{\pi x}2(-3)+(4c-3x)\frac\pi2=\frac\pi2(-3x+4c-3x)=\pi(2c-3x).$$

At $x=4c/3$, $V''=\pi[2c-3(4c/3)]=-2c\pi<0$, since $c>0$. Volume is maximum at $x=4c/3$.

$$V_{\max}=\frac\pi2\left(\frac{4c}3\right)^2\left(2c-\frac{4c}3\right)=\pi\left(\frac{4c}3\right)^2\left(c-\frac{2c}3\right)=\frac{16\pi c^3}{27}.$$

Surface area $S=\pi x(x+2c)$ is strictly increasing for $x>0$, so largest area corresponds to largest $x$. For a nondegenerate cylinder $0<x<2c$,

$$S=\pi x(x+2c)<\pi(2c)(2c+2c)=8\pi c^2.$$

Thus $S<8\pi c^2$, so it cannot exceed $8\pi c^2$ (shown). If degenerate endpoints are admitted, equality occurs only at $x=2c$, height zero.

Teacher notes

Need to check that $V$ is maximum at $x=4c/3$ using a first or second derivative test.

Answer key

$V=\pi x^2(2c-x)/2$, $S=\pi x(x+2c)$, maximum $V=16\pi c^3/27$.

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