I am lost on this math problem

Question

A rectangular piece of metal is 10 in. longer than it is wide. Squares with sides 2 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 1008 in^3, what were the original dimensions of the piece of? metal?

 

What is the original width ____in

Answer 1

First, let's make some equations:

 

Let W = the width of the piece of metal

Let L = the length of the piece of metal

 

The problem tells us the following:

 

W + 10 = L

 

For the second part of the problem, we need to solve the volume of the box. We know that 2 inches are cut from each side of the box, which means the following:

 

The width of the fold = W - 4

The length of the fold = L - 4

The height of the fold = 2

 

Then, according to the volume formula:

 

length * width * height = 1008

(L-4) * (W-4) * 2 = 1008

 

Since we know that L = 10 + W, just substitute and solve for W:

 

(10 + W - 4) * (W - 4) * 2 = 1008

(6 + W)(W - 4)*2 = 1008

 

Divide both sides by 2:

 

(W + 6)(W - 4) = 504

W² + 2W - 24 = 504

W² + 2W - 528 = 0

 

Using the quadratic equation, the two solutions for W are 22 and -24. Negative width makes no sense, so W = 22.

Since L = 10 + W, L = 32.

 

We can check our solution by plugging the numbers into the volume formula:

 

(L-4) * (W-4) * 2 = 1008

(32-4)(22-4)(2) = 1008

28*18*2 = 1008

1008 = 1008

 

So our answer checks out