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Misha Larkins [42]
3 years ago
13

On a certain​ route, an airline carries 8000 passengers per​ month, each paying ​$50. A market survey indicates that for each​ $

1 increase in the ticket​ price, the airline will lose 100 passengers. Find the ticket price that will maximize the​ airline's monthly revenue for the route. What is the maximum monthly​ revenue?
Mathematics
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

  • $65
  • $424500

Step-by-step explanation:

Let x be the price increase.

<u>Then the current price is: </u>

  • 50 + x

<u>Then the number of passengers is:</u>

  • 8000 − 100x

<u>Then the revenue: </u>

  • Revenue = Price * Number of Passengers
  • R = (50 + x)(8000 − 100x) =
  •       - 100x² + 3000x+ 400000  

This is a quadratic relationship. Maximum point is obtained at vertex.

<u>Vertex is determined by x = - b/2a, find the value of x:</u>

  • x = - 3000 / (-2*100) = 15

<u>The maximum revenue is:</u>

  • R = (50 + 15)(8000 − 100 * 15) = 424500
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Step-by-step explanation:

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We know that Jayanta can only spend 20 hours letter writing and 14 hour of follow-up.

So, equations becomes:

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And total money raised can be shown by = 150c+175l

We have to maximize 150c+175l keeping in mind that 2c+2l \leq 20 and c+3l \leq 14

We will solve the two equations:  2c+2l=20 and c+3l=14

We get l = 2 and c = 8

And total money raised is 150\times8 + 175\times2 = 1200+350=1550 dollars.

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Step-by-step explanation:

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