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Maximum Revenue When Money, Power Inc. charges $600 for a seminar on management it attracts 1000 people. For each $20 decrease in the lee an additional 100 people will attend the seminar. The mangers wonder how much to charge for the seminar to maximize their revenue 1) Write the revenue function for ticket sales. Let x= the number of $20 decreases to the ticket price 2) Determine the price per ticket that should be charged in order to maximize total revenue.

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Answer:

Step-by-step explanation:

1) To write the revenue function for ticket sales, we need to consider two factors: the ticket price and the number of attendees.

Let's denote the original ticket price as P and the number of $20 decreases in the ticket price as x. Since each decrease in price attracts an additional 100 people, the total number of attendees can be expressed as (1000 + 100x).

The revenue is calculated by multiplying the ticket price by the number of attendees. Therefore, the revenue function for ticket sales can be written as:

Revenue(x) = (P - 20x) * (1000 + 100x)

2) To determine the price per ticket that should be charged to maximize total revenue, we can find the value of x that maximizes the revenue function.

To do this, we can take the derivative of the revenue function with respect to x, set it equal to zero, and solve for x. This will give us the value of x at which the revenue is maximized.

Revenue'(x) = 1000 - 2000x + 2000

0 = 1000 - 2000x + 2000

2000x = 2000

x = 1

Therefore, x = 1 represents the number of $20 decreases in the ticket price that maximizes the revenue.

To determine the price per ticket, we substitute x = 1 into the revenue function:

Revenue(1) = (P - 20 * 1) * (1000 + 100 * 1)

Revenue(1) = (P - 20) * 1100

To maximize total revenue, the price per ticket that should be charged is $20 above the cost, which means:

P - 20 = 600

P = 620

Hence, the price per ticket that should be charged to maximize total revenue is $620.