Hello, PLAYNOISE!
You have 100 meters of fencing material to enclose a rectangular plot.
Your goal is to determine the dimensions of the plot such that you can enclose the maximum area possible.
Solve the problem algebraically by showing that the area function can be written as \(\displaystyle A(x)\:=\:625\,-\,(x\,-\,25)^2\)
How does this form of the function allow you to find the dimensions that produce a maximum area?
I assume you got the first part.
Let \(\displaystyle x\) = length of the plot, \(\displaystyle y\) = width of the plot.
From the fencing restriction: \(\displaystyle 2x\,+\,2y\:=\:100\;\;\Rightarrow\;\;y\:=\:50\,-\,x\)
The area of the plot is: \(\displaystyle \;A\;=\;x\cdot y\;=\;x(50\,-\,x)\;=\;50x\,-\,x^2\)
We want to maximize \(\displaystyle A\).
We note that the area function is a <u>parabola</u>.
. . It reaches is maximum (or minimum) at its vertex.
So they want us to put the equation in "vertex form".
We have: \(\displaystyle y\;=\;50x\,-\,x^2\)
We will complete-the-square: \(\displaystyle \;y\;=\;-(x^2\,-\,50x)\)
Then we have: \(\displaystyle \;y\;=\;-(x^2\,-\,50x\,+\,625\,-\,625)\;=\;-(x^2\,-\,50x\,+625)\,+\,625\)
. . . \(\displaystyle y\;=\;-(x-25)^2\,+\,625\) . . .
There!
. . The vertex is at: \(\displaystyle \;(25,\,625)\)
To maximize the area, use \(\displaystyle x\,=\,25\) m for the length.
. . Then it turns out that \(\displaystyle y\,=\,25\) m is the width.
That is, to maximize the area of the rectangle,
make it a square!