Trial and error?

>>17213659
Like this.

One way could be all four sides are 60 like a square, then 60•60=360

Or try 40/80/40/80. 40•80=320

So right off the bat you might deduce that squares have the highest possible area with that perimeter but not for certain

>>17213651
A = l*w, P = 2*(l+w)

240/2 = l + w, l = 120 - w

A = 120w - w^2, dA/dw = 120 - 2w

When dA/dw = 0, Area is at either max or min, so w = 60, and A = 3600.

Perimeter of a rectangle: 2Y + 2Z = 240
Y = height of one side, Z = width of one side

Y = 60, Z = 60 is a solution to those. So is Y = 80, Z = 40.

If we plug those in, area equals width times height. 80 x 40 = 3200. 60 x 60 = 3600. The max area is 3600.

>>17213674
Then take it one step further.

Get closer to a square then 40/80, with 50/70/50/70. 50•70=350.

So, logic tells you that squares have the highest possible area for rectangles for at least that perimeter if not for all perimetes

>>17213693
>>17213674
3,600, 3,200 and 3,500 sorry

A square is the rectangle with the largest area, always.

A = L x W

So you maximize the area assuming a finite perimeter, you need to bring the length and width to the same value. Since adding anything to width takes away from length and vise versa. E.G.
With perimeters of 20 you can do
9*1=9
8*2=16
7*3=21
6*4=24
5*5=25

How's 5th grade going btw?

>>17213651
Perimeter:
2x+2y = 240
Reduce to
x=120-y
x*y=z
Substitute x from first equation
(120-y)*y=z
120y-y^2=z
First derivative dz/dy is the maximum of z at dz/dy=0
120-2y=0
y=60
sub into first equation, x=60
So max area is 60*60=3600.

>>17213684
Am I the only one who solved the problem the proper way? Everyone else just guessed some values that fit the perimeter or stated the geometrical fact that a square is a solution to this maximisation problem without providing strict mathematical proof.

>>17213728
>>17213698
>>17213693
>>17213674
I did :)