Posted by **kait** on Sunday, September 25, 2011 at 12:26am.

a builder intends to construct a storage shed having a volume of 900ft^3, a flat roof and a rectangular base whose width is three-fourths the length. the cost per square foot of the materials is 4000.00 for the floor,6000.00 for the sides and 3000.00 for the roof. what dimension will minimize the cost?

- diffirential calculus -
**Reiny**, Sunday, September 25, 2011 at 11:18am
I like to avoid fractions if possible, so let the

width be 3x

and the length be 4x , (notice that 3x/4x) = 3/4)

let the height be h

V= (3x)(4x)(h)

h = 900/(12x^2) = 75/x^2

Cost = C = 4000(top) + 6000(sides) + 3000(roof)

= 4000(12x^2) + 6000(8xh+6xh) + 3000(12x^2)

= 84000x^2 + 84000xh

= 84000x^2 + 84000x(75/x^2

= 84000x^2 + 84000(75)/x

= 84000(x^2 + 75/x)

d(Cost)/dx = 84000(2x - 75/x^2) = 0 for a min of C

2x = 75/x^2

x^3 = 75/2 = 37.5

x = (37.5)^(1/3) = 3.347

width = 3(3.347) = 10.04

length = 4(3.347) = 13.39

height = 75/(3.347)^2 = 6.69 (all in feet)

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