A cone-roof model with least surface area · source reference

Printed question

[It is given that a right circular cone with base radius $r$ and height $h$ has volume $\frac13\pi r^2h$ and curved surface area $\pi r\ell$, where $\ell$ is the slanted height of the cone.]

The model of a house is made up of three parts.

• The roof is modelled by the curved surface of a right circular cone with base radius $15x$ cm and height $8x$ cm.
• The walls are modelled by the curved surface of a cylinder of radius $15x$ cm and height $w$ cm.
• The floor is modelled by a circular disc of radius $15x$ cm.

The three parts are joined together as shown in the diagram. The model is made of material of negligible thickness. It is given that the volume of the model is a fixed value $k\,\mathrm{cm}^3$, and the external surface area is a minimum. Use differentiation to find the exact values of $x$ and $w$ in terms of $k$. Simplify your answers. [8]

[2022/NYJC/Prelim/I/Q2]

Printed working

Volume $k=\pi(15x)^2w+\frac13\pi(15x)^2(8x)$ (1).

Surface area:

$$\begin{aligned}A&=2\pi(15x)w+\pi(15x)^2+\pi(15x)\sqrt{(8x)^2+(15x)^2}\\&=2\pi(15x)w+\pi(15x)^2+\pi(15x)(17x)\\&=30\pi xw+480\pi x^2\quad(2).\end{aligned}$$

From (1), $w=(k-600\pi x^3)/(225\pi x^2)$. Substitute into (2):

$$A=30\pi x\left[\frac{k-600\pi x^3}{225\pi x^2}\right]+480\pi x^2=\frac{2k}{15x}+400\pi x^2,$$ $$\frac{dA}{dx}=-\frac{2k}{15x^2}+800\pi x.$$

Let $dA/dx=0$: $x^3=k/(6000\pi)\Rightarrow x=\frac1{10}(k/(6\pi))^{1/3}$.

$$\begin{aligned}w&=\frac1{225\pi\,k^{2/3}/[100(6^{2/3})(\pi^{2/3})]}\left[k-600\pi\frac{k}{6000\pi}\right]\\&=\frac{100(6^{2/3})}{225\pi^{1/3}k^{2/3}}\left[\frac9{10}k\right]\\&=\frac{12}5\left(\frac{k}{6\pi}\right)^{1/3}.\end{aligned}$$

Teacher notes

$k$ is a constant since the volume of the model is fixed.

Express $A$ in terms of only one variable $x$, eliminating $w$ using equations (1) and (2).

Given that $A$ is a minimum, set $dA/dx=0$.

Answer key

$x=\frac1{10}(k/(6\pi))^{1/3}$; $w=\frac{12}5(k/(6\pi))^{1/3}$.

Native diagrams