You must construct a cylindrical container with no lid. It must hold 200 cm^3. The costof material for the bottem is 0.20 $/cm^2, and cost for material for the side is 0.10$/ cm^2. What are best dimensions to cut cost? Width can not exceed 4 cm.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

This can be doen by using differentiation.

Let the radius of the cylinder be r and the height be h.

Then the total volume is `V = pir^2h`

but we know V = 200 cm3

therefore, h is

`h = 200/(pir^2)`


Now we will get an expression for area,

Side area A1 = = `2pirh `

A1 = `2pir*(200/(pir^2))`

A1 = `400/r`

The cost for this side area C1 = 0.1 * A1

C1 = `0.1*(400/r)`

C1 = `40/r`


Now the bottom are = A2

A2 = `pir^2`

the cost for bottoma area = C2 = `0.2*A2`

C2 = `0.2*pir^2`

C2 = `0.2pir^2`


Now th total cost will be C = C1+C2

`C = 40/r + 0.2pir^2`


we have to find the r value which minimise the cost. So we have to take the derivative of C wrt r

`(dC)/(dr) = (-40)/r^2+0.4pir`

For critical points, this derivative must be zero, therefore,

` ` `(-40)/r^2+0.4pir = 0`

`r^3 = 100`

`r = 100^(1/3) = 4.6415 cm`


To find whether this is a minimum value, we have to check for the second derivative of C,

`(d^2C)/(dr^2) = 80/r^3 + 0.4pi`

This gives that the second derivative is positive at any value of r, tehrefore at `root(3)(100)` it is positive and C has a minimum at that value.

now h would be,

`h = 200/(pir^2) = 200/(pi(root(3)(100))^2)`

`h = 2.954 cm`

Therfore the dimensions are r = 4.6415 cm and h = 2.954 cm


See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial