Question 1048929: A rain gutter is made from sheets of aluminum that are 17 inches wide. As shown in the figure, the edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 18 square inches. Show that there are two different solutions to the problem. Round to the nearest tenth of an inch.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A rain gutter is made from sheets of aluminum that are 17 inches wide.
As shown in the figure, the edges are turned up to form right angles.
Determine the depth of the gutter that will allow a cross-sectional area of 18 square inches.
Show that there are two different solutions to the problem.
Round to the nearest tenth of an inch.
:
The end of the gutter can be illustrated like this
d|_w_|d
where
d = the depth of the gutter
w = the width
:
then
2d + w = 17 in
w = -2d + 17
and
d * w = 18 sq/in
replace w with (-2d+17)
d(-2d+17) = 18
-2d^2 + 17d = 18
A quadratic equation
-2d^2 + 17d - 18 = 0
Using the quadratic formula, a=-2; b=17 c=-18
Two solutions
d = 7.3" then w = -2(7.3)+17 = 2.4" is the width
and
d = 1.2" then w = -2(1.2)+17 = 14.6" is the width
:
;
Check these by finding the area
7.3*2.4 = 17.52 sq/in, Not 18 because of rounding to the nearest 10th
1.2*14.6 = 17.52
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