Question 149541:  A rectangular garden has dimensions of 18 feet by 13 feet.  A gravel path of uniform width is to be built around the garden.  How wide can the path be if there is enough gravel for 516 square feet? 
 
 Answer by ankor@dixie-net.com(22740)      (Show Source): 
You can  put this solution on YOUR website! A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet? 
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Draw a rectangle representing the 18 by 13 ft garden, then draw a larger rectangle 
around that one, enclosing the path around the garden. 
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Label the width of the path as x 
It will be apparent that the overall dimensions will be (18+2x) by (13+2x) 
FOIL this to get the overall area: 
(18+2x)*(13+2x) = 234 + 36x + 26x + 4x^2 = 234 + 62x + 4x^2 
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Garden area; 18 * 13 = 234 
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The equation: 
Overall area - garden area = path area (given as 516 sq/ft) 
(4x^2 + 62x + 234) - 234 = 516 
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4x^2 + 62x - 516 = 0; our old friend, the quadratic equation 
Simply divide equation by 2: 
2x^2 + 31x - 258 = 0 
Factor this to: 
(2x + 43)(x - 6) = 0 
Positive solution 
x = +6 ft is the width of the path 
: 
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We can check this: overall dimensions will be (18+12) by (13+12) 
(30*25) - 234 = 
750 - 234 = 516, confirms our solution 
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Did this make sense? Any questions? 
 
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