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.

Already a member? Log in here.