An open box is to be constructed so that the length of the base is 4 times large

Question

An open box is to be constructed so that the length of the base is 4 times larger than the width of the base. If the cost to construct the base is 2 dollars per square foot and the cost to construct the four sides is 1 dollars per square foot, determine the dimensions for a box to have volume = 70 cubic feet which would minimize the cost of construction. The values for the dimension of the base are: ?, ? The height of the box is: ?

Explanation / Answer

An open box is to be constructed so that the length of the base is 5 times larger than the width of the base. If the cost to construct the base is 5 dollars per square foot and the cost to construct the four sides is 2 dollars per square foot, determine the dimensions for a box to have volume = 86 cubic feet which would minimize the cost of construction. L = length W = width H = height L = 5W Volume = L*W*H = 86 >>> H = 86 / LW >>> H = 86 / 5W^2 Area of base = L*W Area of 4 sides = 2*H*L + 2*H*W Cost = 5(L*W) + 2(2*H*L + 2*H*W) I want to get the Cost formula in terms of just one variable, let's say W Cost = 5(5W^2) + 2(2*(86/W) + 2*(86/5W)) Cost = 25W^2 + 344/W + 68.8/W Cost = 25W^2 + 412.8/W Take the derivative of this and set it equal to zero and solve for W 50W - 412.8/W^2 = 0 50W^3 - 412.8 = 0 50W^3 = 412.8 W^3 = 8.256 W = 2.02 When the width is 2.02 the cost is minimized. All the other dimensions fall into place: L = 10.10 H = 4.22 So the base is 2.02 x 10.10 and the height is 4.22 An open box is to be constructed so that the length of the base is 5 times larger than the width of the base. If the cost to construct the base is 3 dollars per square foot and the cost to construct the four sides is 3 dollars per square foot, determine the dimensions for a box to have volume = 25 cubic feet which would minimize the cost of construction. L=5 * 60^(1/3) W = 60^(1/3) H = 5 / 60^(2/3) total cost = cost of base + cost of sides = (w*5*w + 2*w*h + 2*5w*h)*3 c = (5ww+12wh)*3 since the volume is 25 ft cube, w*5w*h=25===> wh= 5/w therefore c= (15w^2 + 180/w ) dc/dw = 30w -180/w^2 for optimized width 0= 30w - 180/w^2 w = cube root of 60 l= 5* cube root of 60 h= 25/ 5* cube root of 60* cube root of 60 = 5/ 60 ^ (2/3)

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