A rectangular garden has dimensions of 15 feet by 11 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 192 square feet?
Solution:
Let x = the width of the path
then
(2x+15) by (2x+11) = overall area of the garden and the path
and FOIL this
Overall Area = 4x^2 + 52x + 165
:
Garden area: 15 * 11 = 165
:
Overall area - garden area = path area (given as 192 sq/ft)
(4x^2 + 52x + 165) - 165 = 192
4x^2 + 52x + 165 - 165 - 192 = 0
4x^2 + 52x - 192 = 0
Simplify, divide by 4
x^2 + 13x - 48
Factors to:
(x + 16)(x - 3) = 0
Positive solution
x = 3 ft is the width of the garden path
Check
(21*17) - (15*11) =
357 - 165 = 192
Solution:
Let x = the width of the path
then
(2x+15) by (2x+11) = overall area of the garden and the path
and FOIL this
Overall Area = 4x^2 + 52x + 165
:
Garden area: 15 * 11 = 165
:
Overall area - garden area = path area (given as 192 sq/ft)
(4x^2 + 52x + 165) - 165 = 192
4x^2 + 52x + 165 - 165 - 192 = 0
4x^2 + 52x - 192 = 0
Simplify, divide by 4
x^2 + 13x - 48
Factors to:
(x + 16)(x - 3) = 0
Positive solution
x = 3 ft is the width of the garden path
Check
(21*17) - (15*11) =
357 - 165 = 192
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