Find the answer to the math application problem in the senior high school entrance examination

Solution: (1) let y=kx+b, it is obvious from the image that b=800, and the point (100, 1600),

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 1600x+2400000，

∴w=-24(x-450)

+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.

=85000,

Solve m

=60+ 10

Root number 2≈74, m

=60- 10

Root number 2≈46,

∫0 < m≤70，

∴m≈46.