Question

A piece of cardboard is 2.8 times as long as it is wide. It is to be made into a box with an open top by cutting 2-inch squares from each corner and folding up the sides. Let x represent the width (in inches) of the original piece of cardboard. Answer the following questions.

a) Represent the cardboard in terms of x.

b) Give restrictions on x. What will be the dimensions of the bottom rectangular base of the box?

c) Determine a function V that represents the volume of the box in terms of x.

d) For what dimensions of the bottom of the box will the volume be 520in^3?

e) Find the values of x if such a box has a volume between 600 and 800in^3 will be in between these solutions.

Answer 1

Explanation:

Let x represent the width (in inches) of the original piece of cardboard.

a) Its length is 2.8 times its width so length is 2.8x.

b) Two 2-inch squares will be cut along the width. So x > 2+2 = 4 inches

Width of bottom rectangular base = x - 4

Length of bottom rectangular base = 2.8x - 4

c) Volume of the box, V = Length * Width * Height

Height is 2 inches.

So V = (2.8x - 4)*(x - 4)*2

= 5.6x^2 - 30.4x + 32

d) V = 5.6x^2 - 30.4x + 32 = 520

5.6x^2 - 30.4x - 488 = 0

Using the quadratic equation to solve for x.

x = 12.44 or -7.01

As x cannot be negative, the width is 12.44 inches.

e) Similarly using the quadratic equation and rejecting the negative solutions, the inequality 600 < V < 800 gives the range of x as

13.15 inches < x < 14.76 inches