Question

Rectangle R has varying length l and width w but a constant perimeter of 4 ft.
For what values of l and w will the area of R be greatest? Give an algebraic argument. Give a geometric arguement

Answer 1

Let W = width
Let L = length
The perimeter, P, is a constant, but L and W are varables.
P = 2L + 2W
A = LW
Since we want the area to be the greatest, we write the area in terms of W and L, our variables.
P = 2L + 2W
2L = P - 2W
L = P/2 - W
Substitute L into A:
A = LW
A = (P/2 - W)W
A = WP/2 - W^2
Now you have the area expressed as a function of the width and the perimeter. The primeter is just a constant, but the width is a variable. A function of the form
y = -x^2 - kx (where k is just a number) is a parabola that opens downward. Therefore it has a maximum value.
Our expression for the area of the rectangle is
A = -W^2 - WP/2 which is of the same form. We see from this that its graph is a parabloa that opens downward. The maximum value corresponds to the maximum area of the rectangle.

Answer 2

Is this the answer or do I need to name values?

Answer 3

This is not the answer yet, but there won't be values because you don't have a value for the perimeter.

Answer 4

Well, there was another question before this because this was a two part question but I didn't undertsand this part

Answer 5

@mathstudent55

Answer 6

The last step here on the algebraic side is to find what value of W will produce the greatest area. That happens at W = -P/4.
Now you need to plug in -P/4 into the Area equation to get the maximum area.

Answer 7

Okay, would it be \[A=\frac{ 3P ^{2} }{ 16 } ?\]

Answer 8

I got A = P^2/16

Answer 9

Do you want to see what I did?