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:

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

Step-by-step explanation:

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

[tex]V = l^{2}\cdot h[/tex] (1)

Where:

[tex]l[/tex] - Side length of the box, in feet.

[tex]h[/tex] - Height of the box, in feet.

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

[tex]C = (c_{b}+c_{t})\cdot l^{2} + 4\cdot c_{s}\cdot l\cdot h[/tex] (2)

Where:

[tex]c_{b}[/tex] - Unit cost of the base of the box, in monetary units per square foot.

[tex]c_{t}[/tex] - Unit cost of the top of the box, in monetary units per square foot.

[tex]c_{s}[/tex] - Unit cost of the side of the box, in monetary units per square foot.

By (1), we clear [tex]h[/tex] into the expression:

[tex]h = \frac{V}{l^{2}}[/tex]

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

[tex]C = (c_{b}+c_{t})\cdot l^{2}+4\cdot c_{s}\cdot \left(\frac{V}{l} \right)[/tex] (3)

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

[tex]C = 0.5\cdot l^{2} + \frac{8}{l}[/tex]

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

[tex]C' = l -\frac{8}{l^{2}}[/tex] (4)

[tex]C'' = 1 + \frac{16}{l^{3}}[/tex] (5)

After equalizing (4) to zero, we solve for [tex]l[/tex]: (First Derivative Test)

[tex]l-\frac{8}{l^{2}} = 0[/tex]

[tex]l^{3}-8 = 0[/tex]

[tex]l = 2\,ft[/tex]

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

[tex]C'' = 3[/tex]

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

[tex]h = 5\,ft[/tex]

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

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