Substituting (0,6), (100, 16) into y 1=kx+b, we get
b=6 100k+b= 16，
The solution is k = 0. 1b = 6,
∴y 1=0. 1x+6(x≥0 integer),
Substitute (100, 12) into y2=k 1x,
Solution: k 1=0. 12,
∴y2=0. 12x(x≥0 integer);
∴y 1=0. 1x+6(x≥0 integer), y2=0. 12x(x≥0 integer).
(2) From the meaning of the question, you can get
When y 1 > y2, 0. 1x+6 > 0. 12x, x < 300
When y1= y2,0.1x+6 = 0.12x, x = 300.
When y 1 < y2, 0. 1x+6 < 0. 12x, x > 300;
When x is in the range of 320 ~ 350, it is more cost-effective to choose the first method.