Question

"The cost in dollars of operating a jet-powered commercial airplane Co is given by the following equation
Co = k*n*v^(3/2)
where
n is the trip length in miles,
v is the velocity in miles per hour, and
k is a constant of proportionality.
It is known that at 573 miles per hour the cost of operation is $400 per mile. The cost of passengers' time in dollars equals $282,000 times the number of hours of travel. The airline company wants to minimize the total cost of a trip which is equal to the cost of operating plus the cost of passengers' time.
At what velocity should the trip be planned to minimize the total cost?
HINT: If you are finding this difficult to solve, arbitrarily choose a number of miles for the trip length, but as you solve it, you should be able to see that the optimal velocity does not depend on the value of n."

Answer 1

The cost function is given by the following equation Co = k*n*v^(3/2)

It is known that at 573 miles per hour the cost of operation is $400 per mile. Hence we have 400*n = k*n*(573)^(3/2)

This gives k = 0.029.

Let the velocity be X. The cost of passengers' time = $282,000*(n/X) because time = distance / speed. The total cost of a trip = cost of operating plus the cost of passengers' time.

C = 282000n/X + 0.029n(X)^(3/2)

Maximize the Cost

dC/dX = 0

282000n/X^2 = 0.029n*(3/2)*(X^1/2)

This gives an optimum speed at 529.33 miles per hour.