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Image for 1) An airline requires that the total outside dimensions (length + width + height) of a checked bag not exceed

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Answer #1

l + w + h <= 56

h = w

l + 2h <= 56

For largest volume, l + 2h = 56 --> L = 56 - 2h

V = lwh

V = (56 - 2h)(h)(h)

V = -2h^3 + 56h^2

dV/dh = -6h^2 + 112h = 0

-6h^2 + 112h = 0

h = 112/6

So, h = 56/3 inches

Therefore, w = 56/3 inches

l = 56 - 2h

l = 56 - 2(56/3)

l = 56 - 112/3

l = 56/3 inches

So, the dimensions are :
Length = 56/3 inches
Width = 56/3 inches
Height = 56/3 inches

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C(x) = 16900 + 100x + 0.01x^2

So, average cost = Ca(x) = (16900 + 100x + 0.01x^2) / x

Ca(x) = 16900/x + 100 + 0.01x

dCa/dx = -16900/x^2 + 0.01 = 0

16900/x^2 = 0.01

x^2 = 16900/0.01

x^2 = 1690000

x = 1300 ----> ANSWER

Average cost = 16900/1300 + 100 + 0.01(1300)

Resulting average cost = $126 ---> ANSWER

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3)

L + 2W + 2H <= 126

L + 2W + 2H = 126

If front face is square, then W = H

L + 4W = 126

L = 126 - 4W

Volume V = LWH

V = (126 - 4w)(w)(w)

V = -4w^3 + 126w^2

dV/dw = -12w^2 + 252w = 0

w = 252/12 ---> w = 21

So, h = 21

L = 126 - 4w --> 126 - 4(21)

L = 42

So, max volume is : LWH --> 42*21*21

V(max) = 18522 in^3 -----> ANSWER

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