SOLUTION: Not quite sure where to put this problem but please help me solve. The perimeter of the den is 88ft. If the area is 480 square feet then what are the dimensions of the rectangul

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Not quite sure where to put this problem but please help me solve. The perimeter of the den is 88ft. If the area is 480 square feet then what are the dimensions of the rectangul      Log On


   

Question 83471: Not quite sure where to put this problem but please help me solve.
The perimeter of the den is 88ft. If the area is 480 square feet then what are the dimensions of the rectangular room?
Answer by dolly(163) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length be x ft and the width be y ft.
The perimeter is 88ft
==> 2(l + w) = 88
==> l + w = 88/2
==> l + w = 44
==> w = 44 - l -------------------(1)
Given area = 480
==> l x w = 480
==> l (44-l) = 480
==> 44l - l^2 = 480
==> - l^2 + 44l - 480 = 0
==> l^2 - 44l + 480 = 0
==> l^2 - 20l - 24l + 480 = 0
==> l(l-20) - 24(l - 20) = 0
==> (l - 20) ( l- 24) = 0
==> l-20 = 0 or l - 24 = 0
==> l = 20 or l = 24
==> w = 24 0r w = 20
Thus the dimensions of the retangular room are 20 ft and 24 ft.