# Rectangular Field Along River

• Jan 21st 2007, 06:09 AM
symmetry
Rectangular Field Along River
Jack has 3000 feet of fencing available to enclose a rectangular field. One side of the field lies along a river, so only 3 sides require fencing.

(a) Express the area A of the rectangle as a function of x, where x is the length of the side parallel to the river.

(b) Graph A = A(x). For what value of x is the area largest?
• Jan 21st 2007, 06:42 AM
Soroban
Hello, symmetry!

Did you make a sketch?

Quote:

Jack has 3000 feet of fencing available to enclose a rectangular field.
One side of the field lies along a river, so only 3 sides require fencing.

(a) Express the area $A$ of the rectangle as a function of $x$,
where $x$ is the length of the side parallel to the river.

(b) Graph $A = A(x)$. .For what value of $x$ is the area largest?

Code:

    ~ + ~ ~ ~ ~ ~ ~ + ~       |            |     y|            |y       |            |       * - - - - - - *             x
The area of the field is: . $A \:=\:xy$ [1]

The total fencing is: $x + 2y$.
Since Jack has 3000 feet of fencing: . $x + 2y \:=\:3000\quad\Rightarrow\quad y \:=\:\frac{3000 - x}{2}$ [2]

Substitute [2] into [1]: . $A \:=\:x\left(\frac{3000 - x}{2}\right)\quad\Rightarrow\quad\boxed{A \:=\:1500x - \frac{1}{2}x^2}$ (a)

If we graph $A \:=\:1500x - \frac{1}{2}x^2$, it looks like this:
Code:

        |         |          *         |      *    :    *         |  *      :      *         | *        :        *         |*          :          *         |          :       --*-----------+-----------*--         |        1500        3000
The maximum $A$ occurs when $x = 1500$.

• Jan 22nd 2007, 02:28 PM
symmetry
ok
Soroban,

This reply is one of the best yet on this site. I love the way you break things down step by step.

Thanks!