The length of a rectangle is 1 foot less than 4 times its width. Find the dimensions of this rectangle if the area is 60 square feet.

Question

The length of a rectangle is 1 foot less than 4 times its width. Find the dimensions of this rectangle if the area is 60 square feet.

Answer 1

The area of any rectangle is length times width.

 

Let width = x

Let length = 4x - 1

Area = 60 ft²

 

 

Write an equation that uses the area and the variables.

 

x(4x - 1) = 60

 

4x² - x = 60

 

 

Subtract 60 on both sides of the equation.

 

4x² - x - 60 = 0

 

 

Factor the left side of the equation using FOIL, if possible.

 

(4x + 15)(x - 4) = 0

 

 

Set the factors equal to zero.

 

4x + 15 = 0              and           x - 4 = 0

 

x = -15/4                  and            x = 4

 

 

We have two solutions.  Since we cannot have a negative value for dimensions, we accept the positive value of x.

 

x = 4

 

 

Finally, we substitute this value of x into the dimensions.

 

width = 4 feet

length = 15 feet

 

 

Answer 2

I misread the initial problem. Thank Michael for noticing.

Below is the corrected version.

 

We have the length is 1 ft less than 4 times the width so

L = 4W-1

 

Area = L(W) = 60 ft²

 

(4W-1)(W) = 60

4W²-W = 60

4W²-W-60=0

 

We can factor this by grouping

Multiply the coefficient of the squared term, 4, by

the constant, -60 giving -240.

Look for factors of -240 that add to the coefficient

of the x term, -1.., (-16)(15) = -240, -16+15=-1

Replace the -W with -16W + 15W fiving

 

4W²-16W+15W-60=0

4W(W-4) + 15(W-4) =0

(4W+15)(W-4) = 0

Either...

4W+15=0    or    W-4=0

4W=-15

W=-15/4      or    W=4

 

Disregard the negative answer so W = 4

L = 4W-1=4(4)-1=15

 

Our dimensions are 15 ft by 4 ft