SOLUTION: A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? Give your answ

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? Give your answ      Log On


   

Question 67543: A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? Give your answer in decimal form, rounded to the nearest thousandth.

Answer by aaaaaaaa(138) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is calculated as length times width.
To calculate the area of the garden less area of x width edge, we just subtract x from both the length and width and calculate the area of the smaller rectangle. In algebrese, that is:
a+=+%2830-x%29%2820-x%29
As we know that equals 400, we have an equation!
%2830-x%29%2820-x%29+=+400
600-30x-20x%2Bx%5E2+=+400
x%5E2-50x%2B600+=+400
x%5E2-50x%2B200+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-50x%2B200+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-50%29%5E2-4%2A1%2A200=1700.

Discriminant d=1700 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--50%2B-sqrt%28+1700+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-50%29%2Bsqrt%28+1700+%29%29%2F2%5C1+=+45.6155281280883
x%5B2%5D+=+%28-%28-50%29-sqrt%28+1700+%29%29%2F2%5C1+=+4.3844718719117

Quadratic expression 1x%5E2%2B-50x%2B200 can be factored:
1x%5E2%2B-50x%2B200+=+1%28x-45.6155281280883%29%2A%28x-4.3844718719117%29
Again, the answer is: 45.6155281280883, 4.3844718719117. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-50%2Ax%2B200+%29


As the path cannot be longer than the width of the garden, 20 ft, the width of the path must be approximately 4.38 ft.