Quadratic Equations

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

Thanks, Hectictar.....!!!!