This answer has now been revised. I hope it's correct according to the given figures. Sorry for the error. I must have confused two very similar questions.

The perimeter of the pen is 2L+2W where L and W are length and width.

(a) The fence has a length 2L+W because the shed's length is 30m and forms a width. Since W=30, 2L+30=200, 2L=170, L=85m. The area is 30*85=2550 sq m. [corrected]

(b) The maximum area for the pen is a square, so L=W and the area is L^2. The perimeter is 4L. We know that the shed helps to form a side. So the perimeter is 200+30=230m=4L, so L=57.5m and the area is 3306.25 sq m. [corrected] It's easy to prove that a square is the maximum area for a given perimeter. If L=a-x and W=a+x, the fixed perimeter, P, is 4a, so a=P/4. The area is a^2-x^2. This quantity has a maximum value of a^2 only when x=0, which means L=W=a, a square.

(c) (i) Now, if the length of each pen is L and the total amount of fencing is 200m, then we can write an equation for the total amount of fencing: 3L+3W=200, where W=30, since the shed covers the width of one pen, so L=110/3=36 2/3m. (ii) If the two pens are square for maximum area each, and the side is L, the total enclosed perimeter is 7L. The perimeter is made up of 200m of fence and 30m of shed=230m. L=230/7=32 6/7m=32.857m approx. [corrected]

(d) For three pens of maximum area the perimeter is 10L=200+30=230, and L=23m. The area of each pen is 529 sq m. [corrected] Total pen area is 1587 sq m.