We might as well put the barn at the bottom corner
of the corral.  

We can get the answer very easily, since of all rectangles,
a square has the maximum area for any given perimeter.  The
side of the barn adds 16' to the 280' of fencing, making the 
perimeter 296'.  So since the corral is a square, then each 
side would be 296' divided by 4 or y=74'.  However your teacher
would not accept that reasoning since no algebra is involved.
So I'll do it another way using algebra.



The area of the corral is 

A = y(x+16)

and the fence, adding up the four parts is

(x)+(y)+(x+16)+(y) = 280
        x+y+x+16+y = 280
          2x+2y+16 = 280
             2x+2y = 264  <--divide through by 2
               x+y = 132
                 y = 132-x

So A = y(x+16) become

   A = (132-x)(x+16)
   A = 132x+2112-x2-16x
   A = -x2+116x+2112

This is a parabola that opens down. So its vertex
will be a maximum. The formula for the x-value of the 
vertex is 

x = -b/(2a) = (-116)/[2(-1)] =  58

Therefore y = 132-x = 132-58 = 74, so the corral is
a 74 ft x 74 ft square, as we got without algebra above.

Edwin