Among all the rectangles whose perimeters are 100 ft, find the dimensions of the one with the maximum area.
I'm a little tired......but the area will be maximized when the rectangle is a square with a side of 25
I'll go through it, AT, if you want me to........
Let's call the length of the rectangle L, and the width W .
perimeter = L + L + W + W
perimeter = 2L + 2W
100 = 2L + 2W Solve this for L .
100 - 2w = 2L Divide through by 2 .
50 - W = L
area = L * W Substitute 50 - W in for L .
area = (50 - W) * W
area = 50W - W2
area = -W2 + 50W
The maximum area will be when W = - 50 / [2(-1)] = 25
and L = 50 - W = 50 - 25 = 25
Here it is
The perimeter is given by
P = 2 ( W + L)
P / 2 = W + L ⇒ P/2 - W = L (1)
And the area is
A = L * W sub (1) for L
A = (P/2 - W) * W simplify
(P/2) W - W^2 rearrange
-1W^2 + (P/2)W
a = -1 b = (P/2)
Again....the "W" that maximizes the area is given by -b / [ 2a] =
- ( P/2) / [2 ( -1) ] = - P/ -4 = P /4
And when W = P/4 L = P/2 - P/4 = P/4
So..... W = L = P/4 = 100 / 4 = 25 ft
This will always be true, AT......for any perimeter......area is maximized when W = L = P/4