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 is 400 ft^2, what is the width of the path. I tried (x is w

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> 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 is 400 ft^2, what is the width of the path. I tried (x is w      Log On


   

Question 97846: 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 is 400 ft^2, what is the width of the path.
I tried (x is width of path):
(30-x)(20-x)=400
600-30x-20x+x^2=400
x^2-50x+600=400
x^2-50x+600-400=0
x^2-50x+200=0
I then tried the quadratic equation with a=1, b=-50, c=200 and ended up with x=25 +- 5 radical 17
Help!!!! Thanks!
Found 2 solutions by checkley71, mathslover:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
THIS IS WHY A STRONGLY RECOMMEND YOU DRAW A DIAGRAM BEFORE YOU START TO SOLVE THESE TYPE OF PROBLEMS. YOU SHOULD HAVE DRAWN A 20 BY 30 RECTANGLE.
THIS GIVES YOU A 20*30=600 SQUARE FT.
NOW DRAW A BORDER INSIDE THIS RECTANGLE WITH A STANDARD WIDTH & LABEL THE WIDTH OF THE PATH=X.
NOW SHOULD SEE THAT THE INSIDE AREA THAT=400 SQUARE FT. HAS THE LENGTH OF (30-2X) & HAS A WIDTH OF (20-2X).
NOW SET THE PRODUCT OF THE LENGTH TIMES THE WIDTH EQUAL TO 400.
(30-2X)(20-2X)=400 NOW MULTIPLY THE SIDE DIMENTIONS
600-40X-60X+4X^2=400
4X^2-100X+600-400
4X^2-100+200=0 NOW WXTRACT THE 4 FROM ALL TERMS
4(X^2-25X+50)=0 NOW FACTOR THE EQUATION
USING THE QUADRATIC EQUATION
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(25+-SQRT[-25^2-4*1*50])/281
X=(25+-SQRT[625-200])/2
X=(25+-SQRT425)/2
X=(25+-20.62)/2
X=45.62/2
X=22.81 FT. THIS CANNOT BE THE ANSWER BECAUSE IT IS WIDER THAN THE ENTIRE GARDEN.
X=(25-20.62)/2
X=4.38/2
X=2.19 FT. ANSWER FOR THE WIDTH OF THE PATH.
PLUG IN THESE FIGURES INTO YOUR DIAGRAM FOR THE PROOF.






Answer by mathslover(157) About Me  (Show Source):
You can put this solution on YOUR website!
you were close
the place where you erred was in taking the length as 30 -x
Since its a path running around you have to consider 2 sides.So the length
would be 30 -2x and the width 20 -2x
Now you can procced as earlier,
(30-2x)(20-2x)=400
600-100x + 4x^2 =400
4x^2 -100x +200=0
x^2 -25x + 50 =0
x= (25 +- 5sqrt(17))/2

we take only
(25 - 5sqrt(17))/2 since the other root makes the widt greater than the side
=2.192 ft