Mathematical problems of inverse proportional function in the second day of junior high school

Solution: (1) Because the area of the square OABC is 9.

Therefore: b (3,3)

And b (3,3) in the function y = k/x (k >; 0, x>0), so: k=3×3=9.

(2) Because the point P(m, n) is on the function y=9/x, mn=9, that is, the area of the rectangular PEOF is 9.

The non-overlapping area of rectangular PEOF and square OABC is S=9/2, so the overlapping area is also 9/2.

Therefore, n=3/2 (half the side length of a square OABC) or m=3/2.

When n=3/2, m=6, so: p (6,3/2)

When m=3/2 and n=6, then: p (3/2,6)

(3)n=9/m

When m > 3, n = 9/m < 3, so the overlapping area is 3n=27/m, so S = 9-27/m..

When m=3, S=9.

When m < 3, the overlapping area is 3m, so S=9-3m.

Note: The key is to find the algebraic expression of the overlapping area of rectangular PEOF and square OABC.