Calculus Optimization Problem

Question

An open box is to be constructed so that the length of the base is 3 times larger than the width of the base. If the cost to construct the base is 4 dollars per square foot and the cost to construct the four sides is 2 dollars per square foot, determine the dimensions for a box to have volume = 99 cubic feet which would minimize the cost of construction.

The values for the dimension of the base are: ___ x ____

The height of the box is:

Answer 1

V=99 cubic feet = volume of an open box = Lwh = Length x width x height

h= 99/Lw = 99/3w^2= 33/w^2

L=3w Length of base = 3 times its width

cost of base = $4 per square foot

cost each side = $2 per square foot

cost of box = 4Lw + 2(2Lh) + 2(2wh) = 4Lw + 4Lh + 4wh = 4(Lw+Lh+wh)

= 4(3w^2 + 3w(33/w^2 +w(33/w^2)

= 12w^2 + 396/w + 132/w

= 12w^2 + 528/w

take the derivative and set = 0, solve for w

C' = 24w -528/w^2 = 0

24w^3 = 528

6w^3 = 132

3w^3 = 66

w^3 = 22

w = cube root of 22

w= about 2.8 feet wide

L=3w= about 3(2.8)=8.4 feet Long

h=33/w^2 = about 33/2.8^2 = about 4.21 feet high

Minimum Cost = 4wL + 2Lh +2wh

= 4(3(2.8)^2+ 2(3(2.8)(33/2.8^2) + 2(2.8)(33/2.8^2)

= 94.08 + 198/2.8 + 66/2.8

= 94.08 + 264/2.8

= about $188.37= minimum cost

foot note: there are 3 cube roots

but the other 2 are imaginary

giving the box 3 imaginary dimensions

x^3-22 = (x-22^(1/3))(x^2 +x22^(1/3) + 22^(2/3))

set = factor = 0 and solve for x

x= 22^(1/3) feet is the real solution

and use the quadratic formula on the 2nd quadratic factor

x =-(22^(1/3)/2 +/-(1/2)sqr(22^(2/3) -4(22^(2/3))

=about -1.4 +/- .5isqr(2(22^(2/3)) are two imaginary solutions