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11142009  #1 
Join Date: Nov 2009
Posts: 1

Calculus I optimization word problem
A box is to be made out of a 6 by 18 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length (L) , width (W), and height (H) of the resulting box that maximizes the volume.
(Assume that W is < or = to L). 
11162009  #2  
Join Date: Dec 2008
Posts: 249

Max Volume Box
Quote:
L = length H = height V = volume = W*L*H S = length of the square cutout W = 6  2S L = 18  2S H = S V = (6  2S)(18  2S)(S) V = (6  2S)(18S  2S^2) V = 108S  36S^2  12S^2 + 4S^3 V = 4S^3  48S^2 + 108S To maximize the volume, take the derivative and set it equal to 0. der(V) = 12S^2  96S + 108 0 = 12S^2  96S + 108 0 = 12(S^2  8S + 9) 0 = S^2  8S + 9 Use the quadratic formula to solve for S. S = (8 + sqrt((8)^2  4(1)(9)))/2(1) S = (8 + sqrt(28))/2 S = 4 + sqrt(7) S = 4 + sqrt(7) gives a value greater than 6 which cannot be used since the width is only 6. Therefore: S = 4  sqrt(7) which is about 1.35425 Plug back in to find W, L, H, and V. W = 6  2(1.35425) = 3.2915 L = 18  2(1.35425) = 15.2915 H = 1.35425 Max Volume = 68.16 

02212010  #3 
Join Date: Feb 2010
Posts: 388

Max Volume = 68.16

02222010  #4 
Join Date: Feb 2010
Posts: 388

Max Volume = 68.16

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