SOLUTION: I have one more problem that is stumping me. It reads: An open rectangular gutter is made by turning up the sides of a piece of metal 20 in. wide. The area of the cross-section

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I have one more problem that is stumping me. It reads: An open rectangular gutter is made by turning up the sides of a piece of metal 20 in. wide. The area of the cross-section       Log On


   

Question 127206This question is from textbook Introductory Algebra
: I have one more problem that is stumping me. It reads: An open rectangular gutter is made by turning up the sides of a piece of metal 20 in. wide. The area of the cross-section of the gutter is 50in^2. Find the depth of the gutter. Any help on this one would be greatly appreciated!!! This question is from textbook Introductory Algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let the sides of the rectangular gutter be D (for depth) and the length be L.
From the description we have:
1) 2D+L = 20 and...
2) D*L = 50
Rewrite equation 1) as L = 20-2D and substitute into equation 2).
D%2820-2D%29+=+50 Simplify.
20D-2D%5E2+=+50 Subtract 50 from both sides.
20D-2D%5E2-50+=+0 Divide by -2 to ease calculations.
D%5E2-10D%2B25+=+0 Solve by factoring.
%28D-5%29%28D-5%29+=+0
So, D = 5
The depth is 5 inches and the length is (20-2(5)) = 10 inches.