Answer:Let x = length of the plot (in meters).
Then the two lengths use up 2x meters of the fencing,
. . . leaving 100 - 2x meters for the two widths.
This means each width is (100 - 2x)/2 = 50 - x meters.
The area of a rectangle is: Length x Width
. . . so we have: . A .= .x(50 - x)
We have a <u>parabola</u>: . A .= .- x<sup>2</sup> + 50x</sup>
We know that this parabola opens down (don't we?)
. . . so its vertex must be the highest point.
We're expected to know that the vertex is at: . x = -b/2a
For this problem, a = -1, b = 50, so we have: . x = -50/(2(-1) = 25
We have maximum A when x = 25.
This leaves 50 - x = 25 for the width.
Therefore, for maximum area, make the plot a 25-by-25 square.
Step-by-step explanation:
