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07052013  #1 
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? 
07082013  #2  
Join Date: Dec 2008
Posts: 249

Quote:
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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