SOLUTION: The width of a rectangle is 18 feet less than the perimeter. the area of the rectangle is 2,040 square feet. What are the dimensions of the rectangle?

Algebra ->  Expressions-with-variables -> SOLUTION: The width of a rectangle is 18 feet less than the perimeter. the area of the rectangle is 2,040 square feet. What are the dimensions of the rectangle?      Log On


   

Question 737462: The width of a rectangle is 18 feet less than the perimeter. the area of the rectangle is 2,040 square feet. What are the dimensions of the rectangle?
Found 2 solutions by lynnlo, ikleyn:
Answer by lynnlo(4176) About Me  (Show Source):
Answer by ikleyn(53339) About Me  (Show Source):
You can put this solution on YOUR website!
.
The width of a rectangle is 18 feet less than the perimeter. the area of the rectangle is 2,040 square feet.
What are the dimensions of the rectangle?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

W = (2W + 2L) - 18

Hence  W = 18 - 2L.


Write an equation for the area of the rectangle

    LW = 2040  square feet

    L*(18-2L) = 2040

    18L - 2L^2 = 2040

    2L^2 - 18L + 2040 = 0

    L^2 - 9L + 1020 = 0


Look at the discriminant

    d = b^2 - 4ac = (-9)^2 - 4*1*1020 = 81 - 4080 is negative number.


It tells that this quadratic equation has no real solutions.


Hence, the problem is defective and describes a situation which never may happen in real world.

Solved, with explanations.