SOLUTION: I have tried several times to work this problem and cannot seem to get the correct answer. Any help would be great! Thanks! A rectangular garden has dimensions of 17 feet by 1

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have tried several times to work this problem and cannot seem to get the correct answer. Any help would be great! Thanks! A rectangular garden has dimensions of 17 feet by 1      Log On


   

Question 189231: I have tried several times to work this problem and cannot seem to get the correct answer. Any help would be great! Thanks!
A rectangular garden has dimensions of 17 feet by 11 feet. A gravel path of equal width is to be built around the garden. How wide can the path be if there is enough gravel for 288 square feet?
Answer by jonvaliente(64) About Me  (Show Source):
You can put this solution on YOUR website!
The area of the garden is given by 17 x 11 = 187
If x=width of the path then the dimensions of the garden with the path would be:
(17+2x) long by (11+2x) wide (remember that the width of the path would add to all 4 sides)
The whole garden can have the maximum area of 288+187= square feet, so:
(2x+17)*(2x+11)=475
4x%5E2%2B56x%2B187=475
Subtracting 475 from both sides, we get:
4x%5E2%2B56x-288=0
Factoring, we get:
%282x%2B36%29%2A%282x-8%29=0
2x+36=0 and/or 2x-8=0
2x=-36 2x=8
x=-18 x=4
We take 4, since the width of the path cannot be negative.
The width of the path is 4 feet.