Question

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material for the sides costs $0.10/ft2, and the material for the top costs $0.13/ft2, determine the dimensions (in ft) of the box that can be constructed at minimum cost.

a.length ft

b.width ft

c.height ft

Answer 1

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box
`V=x^2z`

Volume of the box
`=12 ft^3\\`

`Therefore, x^2z=12\\z=(12)/(x^2)`

The material for the base costs
`\$0.17/ft^2`, the material for the sides costs
`\$0.10/ft^2`, and the material for the top costs
`\$0.13/ft^2`.

Area of the base
`=x^2`

Cost of the Base
`=\$0.17x^2`

Area of the sides
`=4xz`

Cost of the sides=
`=\$0.10(4xz)`

Area of the Top
`=x^2`

Cost of the Base
`=\$0.13x^2`

Total Cost,
`C(x,z) =0.17x^2+0.13x^2+0.10(4xz)`

Substituting
`z=(12)/(x^2)`

`C(x) =0.17x^2+0.13x^2+0.10(4x)((12)/(x^2))\\C(x)=0.3x^2+(4.8)/(x) \\C(x)=(0.3x^3+4.8)/(x)`

To minimize C(x), we solve for the derivative and obtain its critical point

`C'(x)=(0.6x^3-4.8)/(x^2)\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8/ 0.6\\x^3=8\\x=\sqrt[3]{8}=2`

Recall:
`z=(12)/(x^2)=(12)/(2^2)=3\\`

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet