Mathematics, 23.11.2019 05:31 rcmhargoux9546

Arectangular box is to have a square base and a volume of 20 ft3. if the material for the base costs $0.35 per square foot, the material for the sides costs $0.10 per square foot, and the material for the top costs $0.15 per square foot, determine the dimensions of the box that can be constructed at minimum cost.
a. length ft
b. width ft
c. height ft.

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Answer from: ashley968

The volume of a box is the amount of space in it.

The dimensions that minimize cost are:

$\mathbf{Length = 2ft}$

$\mathbf{Width = 2ft}$

$\mathbf{Height= 5ft}$

The volume of the box is:

$\mathbf{V = 20}$

Assume the dimension of the base is x, and the height is h.

The volume of the box will be:

$\mathbf{V = x^2h}$

So, we have:

$\mathbf{x^2h = 20}$

The area of the sides is:

$\mathbf{A_1 = 4xh}$

The cost of the side material is 0.10.

So, the cost is:

$\mathbf{C_1 = 0.10 \times 4xh}$

$\mathbf{C_1 = 0.4xh}$

The area of the base is:

$\mathbf{A_2 = x^2}$

The cost of the base material is 0.35.

So, the cost is:

$\mathbf{C_2 = 0.35x^2}$

The area of the top is:

$\mathbf{A_3 = x^2}$

The cost of the top material is 0.15.

So, the cost is:

$\mathbf{C_3 = 0.15x^2}$

So, the total cost is:

$\mathbf{C = 0.4xh + 0.35x^2 + 0.15x^2}$

$\mathbf{C = 0.4xh + 0.5x^2}$

Recall that: $\mathbf{x^2h = 20}$

Make h the subject

$\mathbf{h = \frac{20}{x^2}}$

Substitute $\mathbf{h = \frac{20}{x^2}}$ in $\mathbf{C = 0.4xh + 0.5x^2}$

$\mathbf{C = 0.4x \times \frac{20}{x^2} + 0.5x^2}$

$\mathbf{C = \frac{8}{x} + 0.5x^2}$

Differentiate

$\mathbf{C' = -\frac{8}{x^2} + x}$

Set to 0

$\mathbf{-\frac{8}{x^2} + x = 0}$

Rewrite as:

$\mathbf{x = \frac{8}{x^2}}$

Multiply both sides by x^2

$\mathbf{x^3 = 8}$

Take cube roots of both sides

$\mathbf{x = 2}$

Recall that:

$\mathbf{h = \frac{20}{x^2}}$

So, we have:

$\mathbf{h = \frac{20}{2^2}}$

$\mathbf{h = 5}$

Hence, the dimensions that minimize cost are:

$\mathbf{Length = 2ft}$

$\mathbf{Width = 2ft}$

$\mathbf{Height= 5ft}$

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