Question

The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer 1

Here is the full question

A snack food company wishes to have a cylinder package for it's almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer:

The radius and height are both dimension in the cylinder; in order to minimize the cost

radius = 2.515 cm

height = 18.93 cm

Explanation:

We denote the radius of the cylinder to be = r

and the height of the cylinder = h

The volume of a cylinder is known to be = πr²h

Also, from the question; we are also told that the cylinder contains 120 πcm³

i.e πr²h = 120π

Dividing both sides with π; we have:

r²h = 120

`h = (120)/(r^2)`

The base and sides will cost $.01 per cm² to produce

Total cost of the base and side
`c_1` = 0.01 ( πr² + 2πrh)

but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

i.e

cost of the top cylinder
`c_2` = 0.02 ( πr²)

Overall Total cost =
`c_1 + c_2`

= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

= 0.01 πr² + 0.02 πrh + 0.02 πr²

=
`0.03 \pi r^2 + 0.02 \pi r ((120)/(r^2) )`

=
`0.03 \pi r^2 + 2.4 (\pi)/(r)`

Taking the differentiation to find the radius dimension to minimize cost; we have:

`(dc)/(dr) =0` ⇒
`0.06 \pi r^2 - (2.4 \pi)/(r^2) =0`

`0.06 \pi r^2 = (2.4 \pi)/(r^2)`

`r^4 = (2.4 \pi)/(0.06 \pi)`

`r^4 = 40`

`r = \sqrt[4]{40}`

`r= 2.515` cm

However,
`(d^2c)/(dr^2)=0.12 \pi r + (4.8 \pi)/(r^3)`

`(d^2c)/(dr^2)|__(r= 2.515)} =0.12 \pi (2.515) + (4.8 \pi)/((2.515)^3) >0`

Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

To determine the height ; we have:

`h = (120)/(r^2) \\h = (120)/((2.515)^2)`

`h = (120)/(6.34)`

h = 18.93 cm