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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
avatar+94183 
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You forgot that 3x>y

 

Here is the system of equations

 

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

 

https://www.desmos.com/calculator/gb43eqnnk6

Melody  Apr 14, 2017

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