i) TS=PR=x+150
ii) length of fencing = 500 = QR+RS+TS+PT. RS=PT, the width of the pen.
So 500=x+2RS+x+150. Therefore 2x+2RS=350, x+RS=175 and RS=175-x.
iii) area=length x width=(x+150)(175-x)=26250+25x-x^2.
iv) we can write the area in a different way: 26250-(x^2-25x)=26250+12.5^2-(x^2-25x+12.5^2).
So we get area=26250+12.5^2-(x-12.5)^2; 26406.25-(x-12.5)^2.
The maximum value of the area is 26406.25 sq m when x=12.5m. This is the stationary value. Any other value of x will reduce the area from 26406.25 sq m.
v) 26406.25 sq m is the maximum area, = 162.5^2, length=width, a perfect square.