Find the dimensions of a rectangle with perimeter 112 m whose area is as large as possible. step 1 if a rectangle has dimensions x and y, then we must maximize the area a = xy. since the perimeter is 2x + 2y = 112, then y = 56

Please use the solution below:
Let P = perimeter, A = area
As provided above, a = xy and 
We know that the formula to solve for the perimeter of a rectangle is P = 2x + 2y. Using the given 112m, we can solve the perimeter using the formula:
112 = 2x + 2y
56 = x + y

x = 56-y or y = 56-x
Let's solve the perimeter in terms of y using the formula below:
A = (56-y)(y)
Find the derivative of A = 56-y^2 to get the value of y.
dA/dy = 56-2y = 0
y = 56/2
y = 28
To find X, substitute the value of y in the equation x = 56 - y. 
x = 56 - 28
Therefore, x = 28.
We can conclude that the figure is not a rectangle but a square.

40%=2/550=1/230=3/1056=14/25