CalculusYou 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...
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.
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