Optimization, question 14

Topic: Optimization
Difficulty: Challenging
Suggested time: 8 min
Marks: 4
Theory: UBC Math 100 §8 CLP-1 §3.5Key results

A closed cylindrical can must hold 1000 cm³. Find the radius and height that use the least material, and state the relationship between them.

V = \pi r^{2} h, \qquad A = 2\pi r^{2} + 2\pi r h

Show your setup and your justification, then compare against the model solution.

Your answer

Theory UBC Math 100 §8 CLP-1 §3.5Key results

Worked solution

Strategy

Surface area is the quantity to minimise and volume is the constraint. Eliminate the height, differentiate the resulting single-variable function, and justify the critical point with the second derivative.

Working

$\pi r^{2} h = 1000 \implies h = \frac{1000}{\pi r^{2}}$
Solve the constraint for the variable that is easiest to remove.
$A(r) = 2\pi r^{2} + 2\pi r \cdot \frac{1000}{\pi r^{2}} = 2\pi r^{2} + \frac{2000}{r}$
Substitute. The π and one power of r cancel in the second term.
$A'(r) = 4\pi r - \frac{2000}{r^{2}} = 0$
Differentiate, writing 2000/r as 2000r⁻¹ if that helps.
$4\pi r^{3} = 2000 \implies r^{3} = \frac{500}{\pi} \implies r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \ \text{cm}$
Multiply through by r² and solve.
$A''(r) = 4\pi + \frac{4000}{r^{3}} > 0 \ \text{ for } r > 0$
The second derivative is positive throughout the domain, so this is the minimum.
$h = \frac{1000}{\pi r^{2}} = \frac{2}{r^{2}}\cdot\frac{500}{\pi} = \frac{2r^{3}}{r^{2}} = 2r \approx 10.84 \ \text{cm}$
The cheapest closed can is exactly as tall as it is wide, a result worth remembering.

Answer

r = \sqrt[3]{\frac{500}{\pi}} \approx 5.42 \ \text{cm}, \quad h = 2r \approx 10.84 \ \text{cm}

What earns the marks

Uses the volume constraint to eliminate h, giving A(r) = 2πr² + 2000/r.
Differentiates correctly to A′(r) = 4πr − 2000/r².
Solves r³ = 500/π, giving r ≈ 5.42 cm.
Justifies that this is a minimum, and notes that the optimal height equals the diameter.

Check

Material at r = 5 and r = 6 comes to about 557 cm² and 560 cm², both more than the 554 cm² at r ≈ 5.42 cm.

Where this goes wrong

Minimising the volume instead of the surface area. The volume is fixed by the problem; it is the constraint, not the objective. Identify which quantity is being optimised before differentiating anything.

Leans on

Rearranging formulae, from Functions, graphs and inverses

Try a similar question

A rectangular plot is fenced on three sides, with a straight river forming the fourth side. With 240 m of fencing available, what is the largest area that can be enclosed?

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Optimization practice question 14