 word problems #1 - XP Math - Forums Sign Up FREE! | Sign In | Classroom Setup | Common Core Alignment  XP Math - Forums word problems #1 07-05-2013 #1 shalliko   Join Date: Jul 2013 Posts: 1 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?  07-08-2013   #2
MAS1   Join Date: Dec 2008
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 Originally Posted by shalliko 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. Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
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