Question

A closed rectangle box with a volume of 16 3 ft is made from two kinds of materials. The top and bottom are made of material costing 10 cents/ 2 ft and the sides from material costing 5 cents/ 2 ft . Find the dimensions of the box so that the cost of the materials is minimized. [Hint: The cost of each side is equal to the area of the side times its unit price]

Answer 1

Answer:

The dimensions of the box are;

2 feet, by 2 feet, by 4 feet

Step-by-step explanation:

The given parameters are;

The volume of the closed box = 16 ft³

The material with which the top is made = The material with which the bottom is made

The cost of the material with which the top and bottom is made = 10 cents/(ft.²)

The cost of the material with which the side is made = 5 cents/(ft.²)

Let 'x', 'y', and 'z' represent the length, width, and height of the material, we have;

x·y·z = 16...(1)

The total cost of the box = 2×10×x×y + 2×5×x×z + 2×5×y×z

∴ The total cost of the box = 20·x·y + 10·x·z + 10y·z...(2)

From equation (1)

z = 16/(x·y)

Therefore, we have;

20·x·y + 10·x·z + 10y·z. = 20·x·y + 10·x·16/(x·y) + 10y·16/(x·y)

20·x·y + 160/(y) + 160/(x)

Differentiating, we get;

f'(x) = 20·y - 160/x²

f'(y) = 20·x - 160/y²

20·y - 160/x² = 0

y = 8/x²

20·x - 160/y² = 0

x = 8/y²

∴ x = 8/(8/x²)² = x⁴/8

8 = x⁴/x = x³

x = ∛8 = 2

The length, x = 2 feet

y = 8/x² = 8/2² = 2

The width, y = 2 feet

z = 16/(x·y) = 16/(2 × 2) = 4

The height of the box, z = 4 feet

Therefore, the dimensions of the box are 2 feet, by 2 feet, by 4 feet.