SOLUTION: A trough at the end of a gutter spout is meant to direct water away from a house. The homeowner makes the trough from a rectangular piece of aluminum that is 30 in.long and 14 in.

Question 1070823: A trough at the end of a gutter spout is meant to direct water away from a house. The homeowner makes the trough from a rectangular piece of aluminum that is 30 in.long and 14 in. wide. He makes a fold along the two long sides a distance of X inches from the edge. If he wants the trough to hold 280 in. to the third power of water, how far from the edge should he make the fold?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A trough at the end of a gutter spout is meant to direct water away from a house.
The homeowner makes the trough from a rectangular piece of aluminum that is 30 in.long and 14 in. wide.
He makes a fold along the two long sides a distance of X inches from the edge.
If he wants the trough to hold 280 in. to the third power of water, how far from the edge should he make the fold?
:
let x = the fold distance from the edge
End view is something like this
x|__|x
then
(14-2x) = the width of the trough
:
Volume
x(14-2x)*30 = 280
(14x - 2x^2)*30 = 280
420x - 60x^2 = 280
A quadratic equation
-60x^2 + 420x - 280 = 0
Simplify, divide by -10
6x^2 - 42x + 28 = 0
Using the quadratic formula; a=6; b=-42; c=28
Two solutions
x = 6.2538
and
x = .7462 in, this is the reasonable solution (height of the trough)
:
;
Check this, find the volume using x = .7462
the width of the trough: 14 - 2(.7462) = 12.5076 in wide
Vol = 12.5076 * .7462 * 30
Vol = 279.99 ~ 280
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