1.Y 1 = 8x，Y2 = 4x+ 120 2。 I can't draw it for you. Simply put, it is the hypothesis method. If x= 10, y 1=80, y2= 160, suppose x.3. In fact, it is to find the intersection of y 1 and y2. When x=30x, the cost of the two schemes is the same. When x ≤ 30, the cost of y 1 is low, and when x > 30, the cost of y2 is low. Therefore, the second scheme is cost-effective.