SOLUTION: 1. 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 fo

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: 1. 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 fo      Log On


   

Question 155376: 1. 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 jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
# 1

Let x = width of path

First let's find the area of the garden only: Area of Garden = 18*13=234

So the area of the garden only is 234 square feet.

Now let's draw a picture:

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From the picture, notice that the length of the entire rectangle (including the width of the path) is 18%2B2x (notice there are two "x" lengths per side) and the total width is 13%2B2x .


So the area of the entire enclosure (including the path) is the expression %2818%2B2x%29%2813%2B2x%29


%2818%2B2x%29%2813%2B2x%29-234 Now subtract off the area of the garden (we only want the area of the path)


234%2B36x%2B26x%2B4x%5E2-234 FOIL


4x%5E2%2B62x Combine like terms


So the area of the path only is A=4x%5E2%2B62x


A=4x%5E2%2B62x Start with the area of the path


516=4x%5E2%2B62x Plug in A=516 (which is the area of the path)


0=4x%5E2%2B62x-516 Subtract 516 from both sides


Notice we have a quadratic equation in the form of ax%5E2%2Bbx%2Bc where a=4, b=62, and c=-516


Let's use the quadratic formula to solve for x


x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


x+=+%28-%2862%29+%2B-+sqrt%28+%2862%29%5E2-4%284%29%28-516%29+%29%29%2F%282%284%29%29 Plug in a=4, b=62, and c=-516


x+=+%28-62+%2B-+sqrt%28+3844-4%284%29%28-516%29+%29%29%2F%282%284%29%29 Square 62 to get 3844.


x+=+%28-62+%2B-+sqrt%28+3844--8256+%29%29%2F%282%284%29%29 Multiply 4%284%29%28-516%29 to get -8256


x+=+%28-62+%2B-+sqrt%28+3844%2B8256+%29%29%2F%282%284%29%29 Rewrite sqrt%283844--8256%29 as sqrt%283844%2B8256%29


x+=+%28-62+%2B-+sqrt%28+12100+%29%29%2F%282%284%29%29 Add 3844 to 8256 to get 12100


x+=+%28-62+%2B-+sqrt%28+12100+%29%29%2F%288%29 Multiply 2 and 4 to get 8.


x+=+%28-62+%2B-+110%29%2F%288%29 Take the square root of 12100 to get 110.


x+=+%28-62+%2B+110%29%2F%288%29 or x+=+%28-62+-+110%29%2F%288%29 Break up the expression.


x+=+%2848%29%2F%288%29 or x+=++%28-172%29%2F%288%29 Combine like terms.


x+=+6 or x+=+-43%2F2 Simplify.


So the answers are x+=+6 or x+=+-43%2F2

Since a negative width doesn't make sense, this means that the only solution is x=6


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Answer:

So the width of the path can be up to 6 feet (ie 6 feet is the maximum width)