Question

If the width of a rectangle plot is “x” and the length of the plot is 100-x/2, what is the maximum area of the plot?

Answer 1

1250 units squared

Explanation:

if you're allowed to use a calculator;

Answer 2

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 .= .- x² + 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.

Explanation: