diffirential calculus
posted by kait on .
a builder intends to construct a storage shed having a volume of 900ft^3, a flat roof and a rectangular base whose width is threefourths 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?

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)