For each problem, you must:
1. Define the variables
2. Write the function to be maximized or minimized.
3. Write a system of inequalities.
4. Graph and write the vertices of the feasible region.
5. Solve the problem and answer the question.

UCA is going to host a college basketball tournament. The tickets will be sold in advance for $40 each and on the weekend of the event for $50 each. For the tournament to occur, at least 2,000 of the 8,000 tickets must be sold in advance. How many of each type of ticket should be sold to maximize UCA's revenue?

Question

For each problem, you must:
1. Define the variables
2. Write the function to be maximized or minimized.
3. Write a system of inequalities.
4. Graph and write the vertices of the feasible region.
5. Solve the problem and answer the question.

UCA is going to host a college basketball tournament. The tickets will be sold in advance for $40 each and on the weekend of the event for $50 each. For the tournament to occur, at least 2,000 of the 8,000 tickets must be sold in advance. How many of each type of ticket should be sold to maximize UCA's revenue?

anonymous · Answer

u can take $40 tickets as x and $50 tickets as y

firejay5 · Answer

$40 tickets as for in advance and $50 tickets for the weekend

anonymous · Answer

correct

firejay5 · Answer

or could it be the # of tickets sold

hero · Answer

I will post a solution soon

firejay5 · Answer

why is that

hero · Answer

Because I have to make sure I understand and solve it before explaining to you

firejay5 · Answer

okay @Hero

hero · Answer

Okay, I have it. Sorry, I apologize. I saw this problem earlier, but then my internet went down and I had to spend time troubleshooting it. But I have the solution now.

firejay5 · Answer

Awesome @Hero  I feel sorry for you

hero · Answer

1. Assign Variables
x = no. advanced tickets
y = no. of weekend tickets

2. Set Constraint Function:
x + y = 8000, 2000 ≤ x ≤ 8000 

The 2000 ≤ x ≤ 8000 part means number of advanced tickets must be between or equal to 2000 and 8000)

x + y = 8000

3. Setup Revenue Function:
R = 40x + 50y

4. Using the constraint function substitute one of the variables into the Revenue function so that it has only one variable:
x + y = 8000 
y = 8000 - x
R = 40x  + 50(8000 - x)
R = 40x + 40000 - 50x
R = 40000 - 10x

5. Take the derivative of the Revenue function:
R' = -10

R' = -10 means that the slope of the R function will decrease as x increases. Therefore, in order to Maximize R, x needs to be the lowest possible value.
x = 2000 is the lowest possible value. Therefore y = 6000. 

Thus, the solution is 

x = 2000
y = 6000

Which means, the maximum revenue UCA will receive is $380,000:

R = 40x + 50y
R = 40(2000) + 50(6000)
R = 380000