Explanation Given information: The capacity of barrel is 32 π ft 3 and the cost per square foot of the material for the side of barrel is half that of cost per square foot for top and bottom. Calculation: Assume r (in feet) to be the radius of barrel and h (in feet) be its height. Let the cost of material for the side of barrel be x per square feet. This cost is half of the cost of material for top and bottom. The cost of material for top and bottom is twice that of cost of material for the sides of barrel. So, the cost of material for top and bottom of barrel is 2 x per square feet. Volume of the cylindrical barrel is, V = π r 2 h Substitute 32 π for V in V = π r 2 h . V = π r 2 h 32 π = π r 2 h h = 32 r 2 The area ( A 1 ) of top and bottom of the cylindrical barrel is, A 1 = π r 2 + π r 2 = 2 π r 2 Cost of material ( C 1 ) of constructing this area of top and bottom is, C 1 = A 1 ( 2 x ) = 2 π r 2 ( 2 x ) = 4 π r 2 x The area ( A 2 ) of sides of the cylindrical barrel is, A 2 = 2 π r h Substitute 32 r 2 for h in A 2 = 2 π r h . A 2 = 2 π r ( 32 r 2 ) = 64 π r Cost of material ( C 2 ) of constructing the sides of barrel is, C 2 = A 2 x = 64 π r ( x ) = 64 π x r The total cost of material for constructing the barrel is, C = C 1 + C 2 = 4 π r 2 x + 64 π x r The derivative of the total cost with respect to r is, d C d r = d d r ( 4 π r 2 x + 64 π x r ) = d d r ( 4 π r 2 x + 64 π x r − 1 ) = d d r 4 π r 2 x + d d r 64 π x r − 1 = 4 π x d d r r 2 + 64 π x d d r r − 1 Simplify it further. d C d r = 4 π x ( 2 r 2 − 1 ) + 64 π x ( − 1 r − 1 − 1 ) = 8 π x r − 64 π x r 2 Find all the critical points r such that d C d r = 0

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Chapter 4, Problem 55RE

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The dimensions of the barrel that can be constructed at a minimum cost in terms of material used.

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