SOLUTION: A garden area is 30ft. long and 20ft. wide. A path of uniform width is set around the edge. If the remaining garden is 400ft.^2, what is the width of the path?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A garden area is 30ft. long and 20ft. wide. A path of uniform width is set around the edge. If the remaining garden is 400ft.^2, what is the width of the path?      Log On


   

Question 93440: A garden area is 30ft. long and 20ft. wide. A path of uniform width is set around the edge. If the remaining garden is 400ft.^2, what is the width of the path?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If we draw a picture we get



If we let x be the width of the path, notice there are 2 x-values per side. So we add 2x to 20 to get 20+2x. This is the total length of the side that contains 20 feet.

Also, this means we add 2x to 30 to get 30+2x. This is the total length of the side that contains 30 feet.

So that means the total area can be written as:

A=%2820%2B2x%29%2830%2B2x%29



Now lets find the inner area:
A=20%2A30=600 plug in the garden dimensions
So the inner area is 600 sq ft


Now lets find the total area
inner area+remaining area=600+400=1000


So the total area is 1000 sq ft




1000=%2820%2B2x%29%2830%2B2x%29 Plug in A=1000 (this is the total area)


1000=600%2B100x%2B4x%5E2 Foil

0=600%2B100x%2B4x%5E2-1000 Subtract 1000 from both sides

0=4x%5E2%2B100x-400 Combine like terms



Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 4%2Ax%5E2%2B100%2Ax-400=0 ( notice a=4, b=100, and c=-400)

x+=+%28-100+%2B-+sqrt%28+%28100%29%5E2-4%2A4%2A-400+%29%29%2F%282%2A4%29 Plug in a=4, b=100, and c=-400



x+=+%28-100+%2B-+sqrt%28+10000-4%2A4%2A-400+%29%29%2F%282%2A4%29 Square 100 to get 10000



x+=+%28-100+%2B-+sqrt%28+10000%2B6400+%29%29%2F%282%2A4%29 Multiply -4%2A-400%2A4 to get 6400



x+=+%28-100+%2B-+sqrt%28+16400+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)



x+=+%28-100+%2B-+20%2Asqrt%2841%29%29%2F%282%2A4%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%28-100+%2B-+20%2Asqrt%2841%29%29%2F8 Multiply 2 and 4 to get 8

So now the expression breaks down into two parts

x+=+%28-100+%2B+20%2Asqrt%2841%29%29%2F8 or x+=+%28-100+-+20%2Asqrt%2841%29%29%2F8


Now break up the fraction


x=-100%2F8%2B20%2Asqrt%2841%29%2F8 or x=-100%2F8-20%2Asqrt%2841%29%2F8


Simplify


x=-25+%2F+2%2B5%2Asqrt%2841%29%2F2 or x=-25+%2F+2-5%2Asqrt%2841%29%2F2


So these expressions approximate to

x=3.50781059358212 or x=-28.5078105935821


So our possible solutions are:
x=3.50781059358212 or x=-28.5078105935821


Since a negative length doesn't make sense, our only solution is x=3.50781059358212 which is 3.507 to the nearest thousandth




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Check:

1000=%2820%2B2x%29%2830%2B2x%29 Start with the given area function

1000=%2820%2B2%283.508%29%29%2830%2B2%283.507%29%29 plug in x=3.507

1000=%2820%2B7.016%29%2830%2B7.014%29 multiply

1000=%2827.016%29%2837.014%29 Add

1000=999.970224 Since we rounded, this is as close as it gets. So our answer is verified.