Let the length and width of the rectangular enclosure be x and 32-x yards, as shown in the figure above. If the area of the enclosure is represented by A, then A= x (32-x).

For maximum area, derivative dA/dx should be equal to zero. Thus,

dA/dx= 32-2x=0. Solving for x, we will have x=16.

This means that the enclosure should be a square of side 16 yards.

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