Answer:
$378.75
Explanation:
Data provided:
Capacity = 440 passengers
Operating cost = $4,000 + $70(Number of passengers)
Expected number of passengers = 440 - 0.64T
Ticket price = T
Total operational cost = $4000 + $70( 440-0.64T )
Total operational cost = $34,800 - 44.8T
Thus,
Total revenue = Number of passengers × Ticket price
= (440 - 0.64T)T
= 440T - 0.64T²
also,
Total profit ,P(T) = Total revenue - Total operational cost
P(T) = ( 440T - 0.64T²) - (34,800 - 44.8T)
P(T) = - 0.64T² - 34,800 + 484.8T
Now,
Differentiating with respect to ticket price T
P'(T) = -0.64(2)T - 0 + 484.8(1)
or
P'(T) = - 1.28T + 484.8 ..............(1)
For point of maxima or minima
P'(T) = 0
or
- 1.28T + 484.8 = 0
or
1.28T = 484.8
or
T = $378.75
now,
again differentiating (1) to check for maxima or minima
P''(T)= -1.26(1) + 0
P''(T) = -1.26
Since,
P"(T) < 0
Hence,
T = $378.75 will maximise the profit
