Substitute y=kx+800, 1600=k× 100+800,
Solution: k=8
So y=8x+800,
Similarly, let z=ax+c, and it is obvious from the image that c=3000, point (100,2700),
Substitute z = ax+3000,2700 = a×100+3000,
Solution: a=-3,
z =-3x+3000;
(2)∵w = yz =(8x+800)(-3x+3000)=-24x
2
+2 1600x+2400000,
∴w=-24(x-450)
2
+7260000,
Therefore, the subsidy per mu should be x=450 yuan, and the maximum value of W is 7260000 yuan.
At this time y=8×450+800=4400 mu;
(3) ten thousand mu of vegetable greenhouses have been built, and the original average income per mu is
7260000/4400= 1650 yuan,
The equation is (1650+2000)m-650m-25m.
2
=85000,
Solve m
1
=60+ 10
Root number 2≈74, m
2
=60- 10
Root number 2≈46,
∫0 < m≤70,
∴m≈46.