Now you know the perimeter is 38. So the perimeter of our rectangle is x+x+y+y so

2x+2y=38

2y = 38-2x

y=19-x

Now the area is something you need to maximize, and you do this by taking the derivative of the equation of area and finding where it equals 0, which is essentially finding the absolute max. The area of our rectangle is

x*y = A

where A is the area. Now you need to make this one variable so you can take the derivative, so using y=19-x…

x*(19-x) = A

19x-x^2 = A

Now you take the derivative of A and set it equal to 0 to find the max.

A’ = 19 – 2x

19-2x = 0

19 = 2x

x = 9.5

So you know x, the base, is 9.5. Plus this into your equation to find y

y = 19-x

y=19-9.5

y = 9.5

So x = y = 9.5 It’s a square!

This will always be true for any rectangle of a given perimeter. The maximum area will always be a square.

Perimeter = 38

Perimeter = length + breadth + length + breadth

Area = length * breadth

2 (length + breadth) = 38

length + breadth = 19

Suppose u set length and breadth as 18 and 1, Area will be 18sqmts

If u set length and breadth as 17 and 2, Area will be 34sqmts

Studying the same

U should set length and breadth in the mid level of number 19 to have the largest area.

Thus it will be 10 and 9 and Area will be 90sqmts.

But if u need more precision you can set both sides as 9.5 which will make it a square with even larger area 90.25sqmts.

But remember square is also “a rectangle with equal sides”.

so the largest area would be 90.25 square meters