When 7≤x≤ 10, y = 300.

So you can get y=? 50x+600( 1≤x≤6)300(7≤x≤ 10)；

When 7≤x≤ 10, z2=kx+b,

Substituting points (7,80) and (10,20) gives 7k+b = 80 10k+b = 20.

Solution: k =? 20 b=220

So Z2 =-20x+220;

(2) When 1≤x≤6

w =(-50x+600-200)(20x+40)- 1000-500

=- 1000 x2+6000 x+ 14500

=- 1000(x-3)2+23500，

When x=3, w is the largest, and w is the largest = 23500;

When 7≤x≤ 10

w =(300-200)(-20x+220)- 1000-500

=-2000x+20500

Because -2000 < 0, w decreases with the increase of x.

So when x=7, w gets the maximum value, and w is the maximum value =6500.

To sum up, the profit on the third day was the biggest, with the highest profit of 23,500 yuan.

(3)5 { 20×(300-200)+[450( 1-a %)-200]？ 20( 1+2a %)- 1500 } = 27 100，

Let a%=t,

Finishing: 18t2-t-0.08=0,

△=6.76,

Solution: t=≈ 1+2.636 = 0. 1 or t=≈ 1? 2.6 36 (irrelevant, omitted)