Asked by Sharon
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
Answers
Answered by
Reiny
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?"
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?"
Answered by
Sbrooks
Yes
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.