Water Trough

Question

9. 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 20 inches long and 12 inches wide. He makes a fold along the two sides of distance of x inches from the edge. If he wants the trough to hold 360 inches³ of water, how far from the edge should he make the fold?

Answer 1

Hello Jane,

What you have here is a case of algebra which requires application of the Quadratic formula.

You know the starting dimensions of the sheet of aluminum. 20 in by 12 in. Because the homeowner is making a trough by folding up the aluminum at each side, then one of the starting sides is going to be shorter than it was originally, and the volume of the trough (which you can essentially think of as a cube without a top) can be represented by:

Length x Width x Height (All in Inches)

or

20 in * (12 in - 2X in) * (X in) = 360 in³

 

This can be simplified by dividing both sides by 20 in:

 

20 in * (12 in - 2X in) * (X in) = (360 in³/ 20 in)

 

After dividing both sides by 20 in:

(12 in - 2X in) * X in = 18 in²

 

Distribute X in. across (12 in -2X in):

12X in² - 2X² in² = 18 in²

 

This is where it gets tricky. My first instinct was to attempt to simplify this in some way. This often leads to a lot of frustration. Look carefully at what you have in this. You have an X² term an X term, and a constant value. Therefore, you can rearrange this equation like so:

 

1.Subtract 18 in² from both sides:

12X in² - 2X² in² - 18 in² = 0

 

2. Rearrange the equation:

-2X² in² + 12X in² - 18 in² = 0

 

This for matches the form for a quadratic equation (Ax² + Bx +C = 0)

You can then plug these values into the quadratic formula to solve for X:

 

[-b +/- Sqrt(b² - 4ac)]/2a

 

Because there aren't really good math tools here, I won't work that out for you. But I got an answer of X=3.

 

You can check out this page for more info on the Quadratic Formula and Equation:

http://www.purplemath.com/modules/quadform.htm

 

Hope that helps.

 

 

-Brandon