Question

A company manufactures cylindrical metal drums with open tops with a volume of 2 cubic meters. The metal used to manufacture the drums cost $2 per square meter for the sides and $3 per square meter for the bottom. What should the dimensions be of the drums to minimize the cost of metal in their production? what is the ratio height/ radius?

Answer 1

Answer:

The cost is minimized for

$r = 0.75$  and then $h = 1.76$ and the ratio height / radius = 2.3

Step-by-step explanation:

When you solve these kind of problems you always need to ask yourself what am I minimizing or maximizing, for this case you need to minimize the cost function.

Suppose that the bottom of your drum has a radius $r$ and a height and a height $h$ then the are of the circle at the bottom would be

$A_c = \pi r^2$

And the area of the sides would be given by

$A_s = 2\pi rh$

According to what they say the cost of the sides is 2$ per square meter and the cost of the bottom is 3$ per square meter.

So the cost would be given by this function.

Cost = $4\pi rh + 3\pi r^2$

But they mention that the volume = 2 cubic meters. On the other hand

Volume = $\pi r^2 h$ = 2   if we replace on Cost we get that

$\text{Cost} = \frac{8}{r} + 3\pi r^2$

When you find the derivative of that and equal to zero you will find that the the cost is minimized for

$r = 0.75$  and then $h = 1.76$ and the ratio height / radius = 2.3

Answer 2

The dimensions that minimize the cost of metal are radius of 0.75 m, and a height of 1.13 m, and the height/radius ratio is 1.5

The surface area of an open-top cylinder is:

$\mathbf{A = 2\pi rh + \pi r^2}$

The base costs $3, and the sides cost $2.

So, the cost function is:

$\mathbf{C = 2 \times 2\pi rh + 3 \times \pi r^2}$

$\mathbf{C = 4\pi rh + 3\pi r^2}$

The volume of the cylinder is:

$\mathbf{V = \pi r^2h}$

The volume is 2 cubic meters.

So, we have:

$\mathbf{\pi r^2h = 2}$

Make h the subject

$\mathbf{h = \frac{2}{\pi r^2}}$

Substitute $\mathbf{h = \frac{2}{\pi r^2}}$ in $\mathbf{C = 4\pi rh + 3\pi r^2}$

$\mathbf{C = 4\pi r \times \frac{2}{\pi r^2} + 3\pi r^2}$

$\mathbf{C = 4 \times \frac{2}{r} + 3\pi r^2}$

$\mathbf{C = \frac{8}{r} + 3\pi r^2}$

Differentiate

$\mathbf{C' = -8r^{-2} + 6\pi r}$

Set to 0

$\mathbf{ -8r^{-2} + 6\pi r = 0}$

Rewrite as:

$\mathbf{ -8r^{-2} =- 6\pi r }$

Divide both sides by -r

$\mathbf{ 8r^{-3} =6\pi }$

Divide both sides by 8

$\mathbf{ r^{-3} =0.75\pi }$

Rewrite as:

$\mathbf{ r^3 =\frac{1}{0.75\pi }}$

Solve for r

$\mathbf{ r =0.75}$

Substitute $\mathbf{ r =0.75}$ in $\mathbf{h = \frac{2}{\pi r^2}}$

$\mathbf{h = \frac{2}{\pi \times 0.75^2}}$

$\mathbf{h = 1.13}$

The height/radius ratio is:

$\mathbf{\frac{h}{r} = \frac{1.13}{0.75}}$

$\mathbf{\frac{h}{r} = 1.5}$

Hence, the dimensions that minimize the cost of metal are radius of 0.75 m, and a height of 1.13 m, and the height/radius ratio is 1.5

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