General tickets sell for $6 and adult $9. Sell at least 300 student and at least 500 adult. Costs $1 to advertise for adult and $.50 to advertise student. Have at most $400 for advertising. What is most profit that can be made? algebra inequalities

Answer 1

number of generals (student?) --- x , x ≥ 300
number of adults ----- y , y ≥ 500
ad cost = .5x + 1y , where .5x+y ≤ 400

profit = 6x + 9y - .5x - y
= 5.5x + 8y

The question cannot be answered
e.g. If we allow the min of each
x = 300, y = 500
profit = 5.5(300) + 8(500) = 5650
any increase in either x or y would increase the profit,
so the profit would just keep increasing.
What about maximum capacity of the venue ?

Did you mean
" sell at MOST 300 studens and at MOST 500 adults?"

Answer 2

Yes