In 2005, the monorail charged $3 per ride and had an average ridership of about

Question

In 2005, the monorail charged $3 per ride and had an average ridership of about 28,000/day. In Dec '05, the company raised the fare to $5/ride, and average ridership in '06 plunged to around 19,000/day.a)find a linear demand equation b)find the price the company should have charged to maximize revenue from ridership. What is the corresponding daily revenue? c)The company would have needed $44.9 million in revenues from ridership to break even in '06. WOuld it have been possible to break even in '06 by charging a suitable price? Please show your work

Explanation / Answer

The linear demand equation is of the form p=mq+b, where p is the price and q is the quantity of rides sold. Let p be in dollars. Then observe that 3=m*28000+b and 5=m*19000+b. We can solve for m and b putting the above linear equations into a system. 3=28000m+b 5=19000m+b. Then we can see that -2=9000m, so m=-2/9000=-1/4500. Substituting this into either equation above, we can see that 3=28000(-1/4500)+b=-28000/4500+b. So 3=-56/9+b, or b=3+(56/9)=83/9. Thus, the linear demand equation is p=(-1/4500)q+83/9. This can be expressed in terms of q if necessary. b) Revenue is price times quantity, so R=pq=(-1/4500)q^2+83q/9, a quadratic function (of quantity, in this case). Observe that the parabola opens down/is concave down, so it must have a maximum at the vertex. To find this maximum, use q=-b/2a, where a=-1/4500 and b=83/9 are the coefficients attached to the quadratic and linear term, respectively. It is also possible to take the derivative of the revenue function with respect to quantity and set it to 0, since the coordinate that yields zero derivative for a parabola that opens down has a maximum. Then R'(q)=(-1/2250)q+83/9=0. In either case, q=(-83/9)/(-1/2250)=83*2250/9=83*250=20750. Then p=(-1/4500)(20750)+83/9=83/18, which is approximately 4.61111111, rounded to $4.61. Thus, the daily revenue is R=pq=83/18*20750=861125/9, which is approximately 95680.5555556, rounded to $95680.56. c) Using maximum daily revenue, we can multiply it by 365 to get the revenue for 2006, as follows: 861125/9*365=314310625/9, which is approximately 34923402.7778, rounded to $34923402.78. This is approximately $34 million, which falls short of the $44.9 million required to break even.

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