Question 252661
Let x and y be the sides of the rectangle.
Then, its perimeter 2x+2y = 76
                       2y = 76-2x
                        y = (76-2x)/2
                        y = 38-x    ...(1)
The area of the rectangle is 
                       A = xy
                       A = x(38-x)
                       A = 38x-x^2 
Differentiating with respect to x,
                      dA/dx = 38-2x
For maximum or minimum value of A,
                      dA/dx = 0
                      38-2x = 0   
                         2x = 38
                          x = 38/2
                          x = 19
Sustituting in (1), we get
                          y = 38-19 = 19
Therefore the dimensions are 19,19