Optimization — Question 3

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Question 3

Problem:

A cylindrical can is to be made to hold a volume of 1000cm31000 \, \text{cm}^3. The material for the top and bottom costs twice as much per square centimeter as the material for the side.

  • (a) Find the dimensions (radius and height) of the can that minimize the cost of the material.

  • (b) What is the minimum total cost in terms of material area?

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Original worksheet page 1: question and worked solution for 4-8-003
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Question 3 - Solution

Let: - rr: radius of the cylinder - hh: height of the cylinder

Volume constraint: V=πr2h=1000⇒h=1000πr2V = \pi r^2 h = 1000 \quad \Rightarrow \quad h = \frac{1000}{\pi r^2}

Cost: - Side area = 2πrh2\pi r h, unit cost = 1 - Top and bottom area = 2πr22\pi r^2, unit cost = 2

Total cost CC (proportional to material area): C=2πrh+2⋅2πr2=2πrh+4πr2C = 2\pi r h + 2 \cdot 2\pi r^2 = 2\pi r h + 4\pi r^2

Substitute hh: C(r)=2πr⋅1000πr2+4πr2=2000r+4πr2C(r) = 2\pi r \cdot \frac{1000}{\pi r^2} + 4\pi r^2 = \frac{2000}{r} + 4\pi r^2

Differentiate: C′(r)=−2000r2+8πrC'(r) = -\frac{2000}{r^2} + 8\pi r

Set derivative to 0: −2000r2+8πr=0⇒8πr3=2000⇒r3=250π-\frac{2000}{r^2} + 8\pi r = 0 \Rightarrow 8\pi r^3 = 2000 \Rightarrow r^3 = \frac{250}{\pi}

r=250π3,h=1000πr2r = \sqrt[3]{\frac{250}{\pi}}, \quad h = \frac{1000}{\pi r^2}

(a) Optimal Dimensions: r=250π3,h=1000πr2\boxed{r = \sqrt[3]{\frac{250}{\pi}}, \quad h = \frac{1000}{\pi r^2}}

(b) Minimum Cost (in material area units): C=2000r+4πr2where r=250π3\boxed{C = \frac{2000}{r} + 4\pi r^2} \quad \text{where } r = \sqrt[3]{\frac{250}{\pi}}

Original worksheet page 2: question and worked solution for 4-8-003

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