Maximizing Water Flow

[jchurch]

I have a problem....A rain gutter is made from sheets of alum that are 20 inches wide. The instructions are to figure how to turn up the edges of the gutter to allow for the maximum water flow.

Write an expression for the area of the cross section of the rain gutter in the form ax(Squares) + bx + c and set f(x) eaual to this expression.

I think if you turn up the edges 5 inches on each side you will get the maximum water flow of 50"squared. But not sure how it fits into the equation.

 

[pappus]

I have a problem....A rain gutter is made from sheets of alum that are 20 inches wide. The instructions are to figure how to turn up the edges of the gutter to allow for the maximum water flow.

Write an expression for the area of the cross section of the rain gutter in the form ax(Squares) + bx + c and set f(x) eaual to this expression.

I think if you turn up the edges 5 inches on each side you will get the maximum water flow of 50"squared. But not sure how it fits into the equation.

1. Draw a sketch (see attachment)

2. You want to get a rectangular area which is at maximum size.

3. Let a denotes the area, w and h the width and height of the rectangle. Then you know:

\(\displaystyle a=w\cdot h\)
and
\(\displaystyle w+2h=20~\implies~h=10-\frac12 w\)

4. Plug in this term for h into the equation of the area and you'll get a function:

\(\displaystyle a(w)=w \cdot (10-\frac12 w)\)

\(\displaystyle a(w)= -\frac12 w^2+10w\)

5. You'll get the maximum area if a'(w) = 0.

\(\displaystyle a'(w)=-w+10\)

So w = 10. Consequently h = 5.

 

[soroban]

Hello, jchurch!

Did you make a sketch?

A rain gutter is made from sheets of aluminum that are 20 inches wide.
The instructions are to figure how to turn up the edges of the gutter
. . to allow for the maximum water flow.

Write an expression for the area of the cross-section of the rain gutter
. . in the form \(\displaystyle f(x) \:=\:ax^2+bx+c\)

I assume the edges are turned up perpendicular to the base.
That is, the gutter has a rectangular cross-section.

        *             *
        |             |
      x |             | x
        |             |
        * - - - - - - *
             20-2x

The aluminum is 20 inches wide.
\(\displaystyle x\) inches are turned up on both ends.
Hence, the base is \(\displaystyle (20\!-\!2x)\) inches.

The area of the gutter is: .\(\displaystyle A \:=\:x(20-2x)\)

Therefore: .\(\displaystyle f(x) \:=\:-2x^2 + 20x\)