More Optimization — Question 4

Question 4 - Solution

Let the radius be $r$, and the height be $h$. We are given: $$V = \pi r^2 h = 1000 \Rightarrow h = \frac{1000}{\pi r^2}$$

Surface Area of the Cylinder:

$$A = 2\pi r^2 + 2\pi r h \Rightarrow A(r) = 2\pi r^2 + 2\pi r \cdot \frac{1000}{\pi r^2} = 2\pi r^2 + \frac{2000}{r}$$

Minimize $A(r)$:

Take derivative: $$A'(r) = 4\pi r - \frac{2000}{r^2}$$

Set $A'(r) = 0$: $$4\pi r = \frac{2000}{r^2} \Rightarrow 4\pi r^3 = 2000 \Rightarrow r^3 = \frac{500}{\pi} \Rightarrow r = \sqrt[3]{\frac{500}{\pi}}$$

Now compute $h$: $$h = \frac{1000}{\pi r^2} = \frac{1000}{\pi \left( \sqrt[3]{\frac{500}{\pi}} \right)^2 } = \frac{1000}{\pi} \cdot \left( \frac{\pi}{500} \right)^{2/3} = 1000 \cdot \pi^{-1/3} \cdot 500^{-2/3}$$

Minimum Surface Area:

$$A = 2\pi r^2 + \frac{2000}{r} = 2\pi \left( \sqrt[3]{\frac{500}{\pi}} \right)^2 + \frac{2000}{\sqrt[3]{\frac{500}{\pi}}}$$

$$\boxed{ \begin{aligned} r &= \sqrt[3]{\frac{500}{\pi}} \\ h &= \frac{1000}{\pi r^2} \\ A_{\min} &= 2\pi r^2 + \frac{2000}{r} \end{aligned} }$$