Minimising Surface Area | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Count the surfaces that actually exist

For a cylinder of radius $r$ and height $h$ , volume is circular base area × height: $V=\pi r^2h$ . Surface area depends on whether the ends are closed.

Cylinder formulae are not supplied in MF19, the exam formula booklet. Build them from the circle and rectangle formulae.

Two ends of area πr², plus a side of area 2πrh. An open top removes one end.

The curved side unwraps to a rectangle of width $2\pi r$ , the circumference, and height $h$ . Its area is $2\pi rh$ . Each circular end contributes $\pi r^2$ .

\begin{aligned}A_{\text{closed}}&=2\pi r^2+2\pi rh\\A_{\text{open top}}&=\pi r^2+2\pi rh\end{aligned}

Use the fixed volume to remove the height

Worked example

A closed cylinder has volume 432π cm³. Find the least surface area.

\begin{gathered}\pi r^2h=432\pi\\h=\frac{432}{r^2}\end{gathered}

The height varies with the radius. Substitute it into the area before differentiating:

\begin{gathered}A=2\pi r^2+\frac{864\pi}{r}\\r>0\\A'=4\pi r-\frac{864\pi}{r^2}\\A'=0\\4\pi r^3=864\pi\\r=6\end{gathered}

Multiplying by $r^2$ is valid because $r>0$ . Now $h=12$ cm and $A''=4\pi+1728\pi/r^3>0$ . Also, $A'$ changes from negative to positive at 6 over the whole domain.

Answer

A_{\min}=216\pi\text{ cm}^2

If different surfaces have different prices per unit area, multiply each surface area by its own price before adding. Minimise that cost expression, not the unweighted area.

Change the container, then change the model

Your turnPractice

An open-top cylindrical container has volume $27\pi\text{ cm}^3$ . Find its dimensions for minimum surface area, and that area. Explain what changes from the closed-cylinder model.

Show worked answer

Remove one circular end. The constraint gives $h=27/r^2$ .

\begin{aligned}A&=\pi r^2+\frac{54\pi}{r}\\A'&=2\pi r-\frac{54\pi}{r^2}=0\\&\Rightarrow r^3=27\end{aligned}

Thus $r=h=3$ cm and $A=27\pi\text{ cm}^2$ . The derivative is negative before 3 and positive after it for $r>0$ , so the area is least there.