Find the smallest perimeter and the dimensions for a rectangle with an area of 225 in^2

To solve this, let the dimensions be x and y. We know that the area is 225. So, x * y = 225 

Let the perimeter be denoted as P.

P = 2x + 2y 

sine y = 225/x 

P = 2x + 450/x 

P' = 2 - 450/x^2 

P'' = + 900/x^3 

Put P' = 0 

then x = 15 

at x =15 P'' is positive 

So the minimum perimeter, will be:
P = 2 (15 + 15) = 60 cm

1. 15^2 is 15*15 = 225

2. area of square = L*W and if that is total to 36, it would be 6 because if the L was 6 and the W was 6, the area would be 36

3. 256 divided by 4 is 64ft because if it’s square, all sides are equal. and perimeter is side+side+side+side and if those are equal to 256, your answer is 64