Question

A rectangular box is to have a square base and a volume of 40 ft3. If the material for the base costs $0.34 per square foot, the material for the sides costs $0.05 per square foot, and the material for the top costs $0.16 per square foot, determine the dimensions of the box that can be constructed at minimum cost.

Answer 1

The dimensions of the box so that total costs are minimum are a side length of 2 feet and a height of 5 feet.

Explanation:

Geometrically speaking, the volume of the rectangular box (
`V`), in cubic feet, is represented by this formula:

`V = l^(2)\cdot h` (1)

Where:

`l` - Side length of the box, in feet.

`h` - Height of the box, in feet.

In addition, the total cost of the box (
`C`), in monetary units, is defined by this formula:

`C = (c_(b)+c_(t))\cdot l^(2) + 4\cdot c_(s)\cdot l\cdot h` (2)

Where:

`c_(b)` - Unit cost of the base of the box, in monetary units per square foot.

`c_(t)` - Unit cost of the top of the box, in monetary units per square foot.

`c_(s)` - Unit cost of the side of the box, in monetary units per square foot.

By (1), we clear
`h` into the expression:

`h = (V)/(l^(2))`

And we expand (2) and simplify the resulting expression:

`C = (c_(b)+c_(t))\cdot l^(2)+4\cdot c_(s)\cdot \left((V)/(l) \right)` (3)

If we know that
`c_(b) = 0.34\,(m.u.)/(ft^(2))`,
`c_(s) = 0.05\,(m.u.)/(ft^(2))`,
`c_(t) = 0.16\,(m.u.)/(ft^(2))` and
`V = 40\,ft^(3)`, then we have the resulting expression and find the critical values associated with the side length of the base:

`C = 0.5\cdot l^(2) + (8)/(l)`

The first and second derivatives of this expression are, respectively:

`C' = l -(8)/(l^(2))` (4)

`C'' = 1 + (16)/(l^(3))` (5)

After equalizing (4) to zero, we solve for
`l`: (First Derivative Test)

`l-(8)/(l^(2)) = 0`

`l^(3)-8 = 0`

`l = 2\,ft`

Then, we evaluate (5) at the value calculated above: (Second Derivative Test)

`C'' = 3`

Which means that critical value is associated with minimum possible total costs. By (1) we have the height of the box:

`h = 5\,ft`

The dimensions of the box so that total costs are minimum are a side length of 2 feet and a height of 5 feet.