Question

Your challenge is to create a cylindrical can that minimizes the cost of materials but must hold 100 cubic inches. The top and bottom of the can cost $0.014 per square inch, while the sides cost only $0.007 per square inch. Show how you did it too?

Answer 1

`Radius = 1.997\ in` and
`Height = 7.987\ in`

`Cost = \$1.05`

Explanation:

Given

`Volume = 100in^3`

`Cost =\$0.014` -- Top and Bottom

`Cost =\$0.007` --- Sides

Required

What dimension of the cylinder minimizes the cost

The volume (V) of a cylinder is:

`V = \pi r^2h`

Substitute 100 for V

`100 = \pi r^2h`

Make h the subject

`h = (100 )/(\pi r^2)`

The surface area (A) of a cylinder is:

`A = 2\pi r^2 + 2\pi rh`

Where

`Top\ and\ bottom = 2\pi r^2`

`Sides = 2\pi rh`

So, the cost of the surface area is:

`C = 2\pi r^2 * 0.014+ 2\pi rh * 0.007`

`C = 2\pi r(r * 0.014+ h * 0.007)`

`C = 2\pi r(0.014r+ 0.007h)`

Substitute
`h = (100 )/(\pi r^2)`

`C = 2\pi r(0.014r+ 0.007*(100 )/(\pi r^2))`

`C = 2\pi r(0.014r+ (0.007*100 )/(\pi r^2))`

`C = 2\pi r(0.014r+ (0.7)/(\pi r^2))`

`C = 2\pi (0.014r^2+ (0.7)/(\pi r))`

Open bracket

`C = 2\pi *0.014r^2+ 2\pi *(0.7)/(\pi r)`

`C = 0.028\pi *r^2+ (2\pi *0.7)/(\pi r)`

`C = 0.028\pi *r^2+ (2 *0.7)/(r)`

`C = 0.028\pi *r^2+ (1.4)/(r)`

`C = 0.028\pi r^2+ (1.4)/(r)`

To minimize, we differentiate C w.r.t r and set the result to 0

`C' = 0.056\pi r - (1.4)/(r^2)`

Set to 0

`0 = 0.056\pi r - (1.4)/(r^2)`

Collect Like Terms

`0.056\pi r = (1.4)/(r^2)`

Cross Multiply

`0.056\pi r *r^2= 1.4`

`0.056\pi r^3= 1.4`

Make
`r^3` the subject

`r^3= (1.4)/(0.056\pi )`

`r^3= (1.4)/(0.056 * 3.14)`

`r^3= (1.4)/(0.17584)`

`r^3= 7.96178343949`

Take cube roots of both sides

`r= \sqrt[3]{7.96178343949}`

`r= 1.997`

Recall that:

`h = (100 )/(\pi r^2)`

`h = (100 )/(3.14 * 1.997^2)`

`h = (100 )/(12.52)`

`h = 7.987`

Hence, the dimensions that minimizes the cost are:

`Radius = 1.997\ in` and
`Height = 7.987\ in`

To calculate the cost, we have:

`C = 2\pi r(0.014r+ 0.007h)`

`C = 2* 3.14 * 1.997 * (0.014*1.997+ 0.007*7.987)`

`Cost = \$1.05`