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Old 07-05-2013   #1
shalliko
 
Join Date: Jul 2013
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Default word problems #1

A right cylinder is to be designed to hold 32∏ cubic inches of a soft drink. The cost for the material for the top and bottom of the cylinder is twice the cost of the material of the side. Let r represent the radius of the base and h the height of the cylinder. What are the values of r and h that will produce the minimum manufacturing cost per cylinder?

Can anyone help me with this?
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Old 07-08-2013   #2
MAS1

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Originally Posted by shalliko View Post
A right cylinder is to be designed to hold 32∏ cubic inches of a soft drink. The cost for the material for the top and bottom of the cylinder is twice the cost of the material of the side. Let r represent the radius of the base and h the height of the cylinder. What are the values of r and h that will produce the minimum manufacturing cost per cylinder?

Can anyone help me with this?
This is a calculus problem.

Volume = 32*pi = pi*(r^2)*h so 32 = (r^2)*h

Total Cost = (Area of top)(Cost per unit area of top) + (Area of bottom)(Cost per unit area of bottom) + (Area of side)(Cost per unit area of side)

Since the top and bottom have the same area and are made of the same material then:
Total Cost = 2(Area of top)(Cost per unit area of top) + (Area of side)(Cost per unit area of side)

A1 = area of top
C1 = cost per unit area of top
A2 = area of side
C2 = cost per unit area of side

From the problem statement C1 = 2*C2 so
Total Cost = 2(A1)(C1) + (A2)(C2) = 2(A1)(2*C2) + (A2)(C2)
Total Cost = 4(A1)(C2) + (A2)(C2)

Since the can is a right circular cylinder then:
A1 = pi*r^2
A2 = 2*pi*r*h

Total Cost = 4(pi*r^2)(C2) + (2*pi*r*h)(C2)

Since Volume = 32*pi then 32 = (r^2)*h so r^2 = 32/h and r = 4*sqrt(2/h)
Substituting for r in our total cost equation gives:
Total Cost = (128*pi/h)(C2) + (8*pi*sqrt(2h))(C2)

To minimize the cost we take the derivative of cost in terms of h and then set it equal to 0 and solve for h.

Der(Total Cost) = -128*pi*(C2)/(h^2) + 4*sqrt(2)*pi*(C2)/(sqrt(h)
0 = -128*pi*(C2)/(h^2) + 4*sqrt(2)*pi*(C2)/(sqrt(h)
Multiply both sides by h^2 gives:
0 = -128*pi*(C2) + 4*pi*sqrt(2)(C2)(h^(3/2))
128*pi*(C2) = 4*pi*sqrt(2)(C2)(h^(3/2))
128/(4*sqrt(2)) = h^(3/2)
h^(3/2) = 16*sqrt(2)
Raise both sides of the equation to the 2/3 power giving:
h = (16^(2/3))((2^(1/2))^(2/3)
h = 512^(1/3)
h = 8

Then r^2 = 32/8 = 4, so r = 2.
So when the radius is 2 inches and the height is 8 inches then the cost will be minimized.
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