Solution: This problem is to find the minimum value of open space. If the length is a and the width is 800/A, the open area can be obtained as 3a+a+(800/a-3-1) × 2 = 4a+1600/a-8.

4A+ 1600/A is greater than or equal to twice the square root of < 4a× (1600/a) >, if and only if 4A= 1600/A, take the equal sign (mean value theorem).

Sorry, I can't type the root sign. It's complicated to write. )

When the equal sign is taken, 4A+ 1600/A-8= 152, and the planting area is 648 square meters.

At this time, a is 20m, and 800/a = 40m.

At this time, let A be long. According to the results, A should be 20 meters wide and 40 meters long.

(it should be noted that 20 meters is in front and 40 meters is on both sides, because the initial column questions are done according to the previous A. )