dimensions of a parking lot

Question

a parking lot has a rectangular area of 40,000 yd^2. the length is 200 yd more than twice the width. what are the dimensions of the lot?

Answer 1

So let's call the width x.

The length is 200 more than twice the width can now be represented as 2x + 200.

The area of a rectangle can be found using the formula A = L*W. So we can write

40000 = x ( 2x + 200)

40000 = 2x ^2 + 200x

0 = 2x ^2 + 200x - 40000 

0 = x^2 + 100x - 20000

0 = (x + 200)(x -100)

 

x=-200 or x = 100. A width can't be negative, so x = 100.

 

Check

width is 100

length is 2(100) + 200 = 400

Area is 100*400 = 40000.

 

So the answers work.

The dimensions are 100 by 400.