You need to construct an open-top rectangular box with a base that has a length that is 3.5 times the width and that must hold a volume of exactly 825 cm3. The material for the base of the box costs 8 cents/cm2 and the material for the sides of the box costs 7 cents/cm2.

Find the dimensions for the box that will minimize the cost of the materials used to construct box. (Remember, you are not supposed to use calculus to do this.)

Be sure you define all variables you use and then, graph using your calculator to answer the question.

We want to find dimensions of an open-top box of volume 825 cm³ such that cost of materials is minimized. The length relative to the width is specified, as are the costs of side and bottom materials.

Setup

Variables and dimensions

Let x represent the width of the base of the box. Then the length is 3.5x. The height is found from ...

V = LWH

H = V/(LW) = 825 cm³/(x(3.5x)) = 1650/(7x²)

So, the dimensions of the box are ...

width: x
length: 3.5x
height: 1650/(7x²)

Cost of materials

The cost of materials for the base is $0.08/cm². The area of the base is the product of length and width.

base cost = $0.08 × LW = $0.08 × (3.5x³) = $0.28x²

The cost of materials for the sides is $0.07/cm². The total area of the sides of the box is the product of the height and the perimeter of the base.

side cost = $0.07 × (H)(2(L+W)) = $0.14(1650/(7x²)(x +3.5x) = 148.5/x

The total cost (y) is the sum of the base cost and the side cost. This is the value we want to minimize.

total cost = 0.28x² +148.5/x

Solution

Graph

A graph of the cost versus the width of the box is attached. It shows the minimum cost to be about $34.67 when the width of the box is 6.425 cm.

Dimensions

The remaining dimensions of the box are ...

length = 3.5(6.425 cm) = 22.488 cm

height = 1650/(7·6.425²) cm = 5.710 cm

width = 6.425 cm

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Additional comment

You will find that the cost of the base is half the cost of the sides. This is the general solution to the problem, and can be used to find the exact dimensions without graphing or calculus.

Unfortunately, that fact cannot be shown definitively without using calculus.