Answer to Question #89391 in Calculus for Henry

Question

A blow-moulded polymer container can be considered as a cylinder with flat ends. Its capacity is 1 litre and it has thin walls of uniform thickness.
 Produce expressions for its volume and surface area.
 Using differential calculus, find the dimensions of the cylinder which will result in the minimum amount of polymer being used for its manufacture.

Answer 1

Produce expressions for the volume of the cylinder V and its surface area A. 

Let R be the radius of a cylinder's base , H be the height of the cylinder. Then

$$V=\pi R^2H$$

$$A=2\pi R^2+2\pi RH$$

Using differential calculus, find the dimensions of the cylinder which will result in the minimum amount of polymer being used for its manufacture.

Solve first equation for H

$$H={V \over \pi R^2}$$

Substitute in the equation for A

$$A=A(R)=2\pi R^2+2\pi R{V \over \pi R^2}=2\pi R^2+{2V \over \pi R}, R>0$$

Find the first derivative with respect to R

$$A'(R)=(2\pi R^2+{2V \over \pi R})'=4\pi R-{2V \over \pi R^2}$$

Find the critical number(s)

$$A'(R)=0=>4\pi R-{2V \over \pi R^2}=0$$

$$R=\sqrt[3]{{V \over 2\pi^2}}$$

First Derivative Test

$$0<R<\sqrt[3]{{V \over 2\pi^2}}, A'(R)<0, A(R) \ decreases$$

$$R>\sqrt[3]{{V \over 2\pi^2}}, A'(R)>0, A(R) \ increases$$

The function A(R) has a local minimum at $R=\sqrt[3]{{V \over 2\pi^2}}.$

Since the function A(R) has the only extremum, then A(R) has the absolute minimum at $R=\sqrt[3]{{V \over 2\pi^2}}.$

Then

$$H={V \over \pi (\sqrt[3]{{V \over 2\pi^2}})^2}=\sqrt[3]{{4V \over \pi}}$$

Given that $V=1l=1dm^3.$ Then

$$R=\sqrt[3]{{1 dm^3 \over 2\pi^2}}={\sqrt[3]{4\pi} \over 2\pi}dm\approx0.370 dm=3.70cm$$

$$R=\sqrt[3]{{4(1 dm^3) \over \pi}}={\sqrt[3]{4\pi^2} \over \pi}dm\approx1.084 dm=10.84 cm$$