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# Optimization Problems

An open-top box is to be constructed so that its base is twice as long as it is wide. Its volume is to be 2300 cm^3. FInd the dimensions that will minimize the amount of cardboard required.

Make:
Length = 2w
Width = w
Height = h

### 1 Answer by Expert Tutors

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
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Width = W
Length = 2W
Height = h

Volume = 2300 cm3 = L*W*h = 2W*W*h = 2W2h

So h = 2300/2W2 = 1150/W2

Area = 2*(W*h) + 2*(2W*h) + W*2W = 6900/W + 2W2        [Substituted 1150/W2 for h, L = 2W]

To find the minimum Area, take the derivative of the AREA wrt W, set it to zero, and solve for W:

d(Area)/dW = -6900W-2 + 4W

0 = -6900W-2 + 4W

6900/4 = W3

1725 = W3

12 ≅ W                                               [11.993 rounded to 12]

L = 2W = 24                                      [23.986 rounded to 24]

h = 1150/W2 = 8                              [7.995 rounded to 8]

CHECK:

Volume = 12*24*8 = 2304                 [Within rounding error]