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.

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

The dimensions that minimize cost are:

The volume of the box is:

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

The volume of the box will be:

So, we have:

The area of the sides is:

The cost of the side material is 0.10.

So, the cost is:

The area of the base is:

The cost of the base material is 0.35.

So, the cost is:

The area of the top is:

The cost of the top material is 0.15.

So, the cost is:

So, the total cost is:

Recall that:

Make h the subject

Substitute in

Differentiate

Set to 0

Rewrite as:

Multiply both sides by x^2

Take cube roots of both sides

Recall that:

So, we have:

Hence, the dimensions that minimize cost are:

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Length = 2 ft

Width = 2 ft

Height = 5 ft

Step-by-step explanation:

Let the square base of the box has one side = x ft

Therefore, area of the base = x² ft²

Cost of the material to prepare the base = $0.35 per square feet

Cost to prepare the base = $0.35x²

Let the height of the box = y ft

Then the volume of the box = x²y ft³ = 20

-----(1)

Cost of the material for the sides = $0.10 per square feet

Area of the sides = 4xy

Cost to prepare the sides of the box = $0.10 × 4xy

                                                             = $0.40xy

Cost of the material to prepare the top = $0.15 per square feet

Cost to prepare the top = $0.15x²

Total cost of the box = 0.35x² + 0.40xy + 0.15x²

From equation (1),

Total cost

Now we take the derivative of C with respect to x and equate it to zero,

= 0

x = 2 ft.

From equation (1),

4y = 20

y = 5 ft

Therefore, Length and width of the box should be 2 ft and height of the box should be 5 ft for the minimum cost to construct the rectangular box.

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