Adult ticket costs = $3

Child ticket cost = $12


Adult food = $8

Child food = $4


Adult profit = $5

Child profit = $10


Maximum of 300 tickets

Ticket sales must be at least $900

Food sales must be at least $1000


Cannot be more than 3 children’s tickets per adult ticket sold.

(These are the graphs that I got I feel like I'm missing some)

x>0 (cannot have negative tickets)

y>0 (cannot have negative tickets)

x+y<300 ( maximum of 300 tickets)

3x+12y>900 ( ticket sales)

8x+4y>1000 (food sales)


Optimisation function = Profit = 5x+10y

Find out the maximum profit the theatre could expect. 

Have I missed a grpah line and would I put the food line and ticket line on the same graph? 

Guest Apr 13, 2017

1+0 Answers


You forgot that 3x>y


Here is the system of equations


Can you understand what i have done to determine the maximum profit?



Melody  Apr 14, 2017

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