Number of advance tickets =x
Number of tickets sold at the play=y
Total number of tickets:
At least 300 tickets must be sold in advance:
x`>=` 300 then y`<=` 400
The total proceeds are equal to 20x+30y. Total proceeds need to be maximized. The maximum profit would be when x is the smallest, x=300, so y=700-300=400,
The maximum profit is (20)(300)+(30)(400)=6000+12000=18000
Graph the three lines:
The feasible region is the triangle formed by x=300, the x axis and the y=700-x line.
The vertices are (300,0), (300,400) and (700,0). These poins represent
(300,0): Only 300 tickets are sold in advance, no tickets are sold at the play, then the profit is = 300x20=6000
(700,0): 700 tickets are sold in advance, thus the profit is = 700x20=14000
(300,400): 300 tickets are sold in advance, 400 at the play, then the profit is the highest = 18000
Thus 300 tickets must be sold in advance and 400 at the play in order to maximize the profit.