Steve S. answered • 04/01/14

Tutor

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Tutoring in Precalculus, Trig, and Differential Calculus

x = the square base’s side lengths in inches

B = area of square base = x^2 in in^2

h = height of container in inches

V = volume of container in in^3

V = Bh = 3456 in^3

h = 3456/x^2

T = cost of top and bottom = 2($4/in^2)(x^2 in^2)

S = cost of sides = 4($2/in^2)(xh in^2)

C = T + S

C = 2(4)(x^2) + 4(2)(x)(3456/x^2)

C = 8x^2 + 27648/x

C = 8x^2 + 27648(1/x)

Extrema occur when C’ = 0:

C’ = 0 = 16x – 27648/x^2

0 = 16x – 27648/x^2

0 = x – 1728/x^2

x = 1728/x^2

x^3 = 1728

x = 12 inches

h = 3456/12^2 = 24 inches

A 12”x12”x24” container will minimize cost.

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The minimum cost is:

C = 8(12)^2 + 27648/12 = 1152 + 2304 = $3,456.