Hi, there--

The Problem:
What are the dimensions of a rectangle that has a perimeter of 38 units an an area of 60 square units?

A Solution:
Let W be the width of the rectangle.
Let L be its length.

Now we need to write two equations that use information in the problem about the length and width 
of the rectangle.

{1} The rectangle has a perimeter of 38 units.
Perimeter means the distance around the outside of the rectangle. We calculate the perimeter P by
adding the two widths and two lengths. In an equation, we write this relationship as
2W + 2L = 38

{2} The area of the rectangle is 60 square units.
The formula for the area of a rectangle is length times width. We can express with relationship 
by the equation,
L * W = 60

Now we have a system of two equations with two variables. We will use the substitution method 
to find the length and width.

Divide every term in the first equation by 2.
2W + 2L = 38
W + L = 19

Rewrite this equation in "W=..." form. (Subtract L from both sides.)
W = 19 - L

We see that W and 19-L are equivalent. Make this substitution in the second equation.
L * W = 60
L * (L - 19) = 60

Multiply to clear the parentheses.
L^2 - 19L = 60

Subtract 60 from both sides of the equation.
L^2 - 19L - 60 = 0

We will solve this equation by translating the it to factored form. We need two numbers whose 
product is 60 and whose sum is -19. The numbers are -15 and -4. The equation in factored form is
(L - 15)(L - 4) = 0

The solutions to this equation are 
L - 15 = 0  OR  L - 4 = 0
L = 15  OR L = 4

If L=15, then the W is 4 units since the area is 60 square units and 15*4=60.
By similar reasoning, if L=4, then the width is 15 units.

We need to check that these values for length and with give the correct perimeter. 
15 + 15 + 4 + 4 = 38, so all is well. THe dimensions of the rectangle are 4 units and 15 units.

Feel free to email me if you have questions about the solution.

Ms.Figgy
math.in.the.vortex@gmail.com