Then the length and width of the rectangular fish pond are X-4 and Y-4 respectively.

Judging from the title: (x-4)(y-4)= 1000.

Therefore, y = [1000/(x-4)]+4 = (4x+984)/(x-4).

The area occupied by farmland is xy, which means finding the minimum value of xy.

Substituting y=(4x+984)/(x-4) gives:

xy=(4x^2+984x)/(x-4)

=4x+ 1000x/(x-4)

=4x+ 1000+4000/(x-4)

= 4(x-4)+4000/(x-4)+ 10 16

Because of x>4, 4 (X-4)+4000/(X-4) > = 2 (4 * 4000) 0.5 = 252.98.

That is xy> = 252.98+1016 =1268.98.

When 4(x-4)=4000/(x-4), xy is the smallest.

The solution is x=35.62.

Y=35.62。

When the length and width of the selected farmland are 35.62 meters, the occupied area of farmland can be minimized.