For this question I only need part e and f a The WMA Bus Lin

For this question ,I only need part e and f.

a. The WMA Bus Lines offers sightseeing tours of Washington, D.C. One tour prices at $7 per person, had an average demand of about 1000 customers per week. When the price was lowered to $6, the weekly demand jumped to about 1200 customers. Assuming that the demand equation is linear, find the tour price that should be charged per person to maximize the total revenue each week. b. Suppose that the demand equation for a monopolist is p = 100 - .01x and the cost function is C(x) = 50x + 10,000. Find the value of x that maximizes the profit and determine the corresponding price and total profit for this level of production. c. If e government imposed an excise tax of $10 per unit how would this affect your results in part b? d. In part b, find the production level at which marginal revenue equals marginal cost. e. On a certain route, a regional airline carries 8000 passengers per month, each paying $50. The airline wants to increase the fair. The market research department estimates that for each $1 increase in fare the airline will lose 100 passengers. Determine the price that maximizes the airline\'s revenue. f. The average ticket price for a concert at the opera house was $50. The average attendance was 4000. When the ticket price was raised to $52, attendance declined to an average of 3800 persons per performance. What should the ticket price be to maximize the revenue for the opera house? (Assume a linear demand curve.)

Solution

e) 8000 passengere per month

each pay $50

$1 increase will loose 100 passesngers

Let $x increases then passesngers lost 100x

So, Revenue , R(x) = (50+x)(8000 -100x)

R(x) = - 5000x + 8000x - 100x^2

maximum profit : x = -b/2a = -(8000/2*-100) = 40

So, Price = 50+x = $ 90 maximium reveneu

f) we have two points for price and avg attendance :( 4000, $ 50) and ( 3800 , $52)

slope = (2/-200) = -1/100 $/per person

p(x) =- x/100 + c wher x is avg attendance and we have to find c

50 = -4000/100 + c ; c = 50 + 40 = 90

p = -x/100 + 90

Revenue , R(x) = x(-x/100 +90) = -x^2/100 + 90x

maximize profit : at x= -b/2a = - (90/2*(-1/100)) = 4500avg attendance

price , p = -4500/100 + 90 = $ 45