Let "w" be the width of the field and "b" the breadth. Then the area enclosed is bw. The total fencing used, so far, is b+ w+ b+ w= 2b+ 2w. If the third fence is parallel to the width, and has length w ("w" and "b" are interchangeable so this is no restriction), then the area is still bw but the total fencing used is 2b+ 2w+ w= 2b+ 3w. Since you have 3000 yards of fencing, you can set 2b+ 3w= 30000 and can the solve for b= (3000- 3w)/2= 1500- (3/2)w. Then the total area is still (1500- (3/2)w)(w)= 1500w- (3/2)w^2.

I don't know what methods you have learned to fimd minimum values for a given function, but because this is posted in "algebra", I would recommend "completing the square".