Optimisation: Minimum Surface Area of a Cylinder

Applied max/min questions have two layers: building the function from a constraint, then running the stationary-point routine on it. The “show that” part in (i) hands you the function so that a slip there does not sink part (ii), examiners design it that way, so never skip (i).

A closed cylinder has radius $r$ cm, height $h$ cm and a fixed volume of $1000\ \text{cm}^3$. (i) Show that the total surface area is $S = 2\pi r^2 + \dfrac{2000}{r}$. [3] (ii) Given that $r$ can vary, find the value of $r$ for which $S$ is a minimum, and show that this value gives a minimum. [5]

The working

(i) Use the volume constraint to eliminate $h$. Volume of a cylinder: $\pi r^2 h = 1000 \;\Rightarrow\; h = \frac{1000}{\pi r^2} \quad \text{(M1)}$

Total surface area of a closed cylinder is two circles plus the curved side: $S = 2\pi r^2 + 2\pi r h \quad \text{(M1)}$

Substitute for $h$: $S = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r} \quad \text{(A1, answer given, working must be complete)}$

(ii) Differentiate (write $\frac{2000}{r}$ as $2000r^{-1}$ first): $\frac{dS}{dr} = 4\pi r - \frac{2000}{r^2} \quad \text{(M1)}$

Set equal to zero and solve: $4\pi r = \frac{2000}{r^2} \;\Rightarrow\; r^3 = \frac{2000}{4\pi} = \frac{500}{\pi} \;\Rightarrow\; r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42\ \text{cm} \quad \text{(M1, A1)}$

Prove it is a minimum with the second derivative: $\frac{d^2S}{dr^2} = 4\pi + \frac{4000}{r^3}$

For any $r > 0$ this is positive, so $S$ is a minimum. (M1 for $\frac{d^2S}{dr^2}$, A1 for the reasoned conclusion.)

Where the marks are won and lost

• In a “show that” part the answer is printed, so the marks are entirely for correct, complete working. A single missing line (for example, never stating $S = 2\pi r^2 + 2\pi rh$) forfeits a method mark even if the final line matches.
• The justification in (ii) is a mark on its own. ”$\frac{d^2S}{dr^2} > 0$ hence minimum” earns it; stopping at $r \approx 5.42$ does not.
• Rewriting $\frac{2000}{r}$ as $2000r^{-1}$ before differentiating prevents the most common slip, differentiating a quotient by inspection and getting the sign or power wrong.

Common mistakes

• Using the open-cylinder area ($\pi r^2 + 2\pi rh$) when the cylinder is closed.
• Differentiating $\frac{2000}{r}$ to $+\frac{2000}{r^2}$ (the negative power makes it $-\frac{2000}{r^2}$).
• Finding $r$ but omitting the minimum-justification line.

Full method: Stationary Points notes. Topic home: Calculus pillar.