Profit Equation Question

Question

For a particular flight from Dulles to SF, USAir uses wide-body jets with a capacity of 450 passengers. It costs the airline $2,000 plus $100 per passenger to operate each flight. Through experience USAir has discovered that if a ticket price is $T, then they can expect (450−0.66T) passengers to book the flight. To the nearest $5, for what value of the ticket price, $T, will the airline's profit be maximized?

Answer 1

The profit is the revenue minus the cost P = R - C. The revenue is given as R = Tn, where T is the ticket price (as labeled in the problem) and n is the number of passengers. The cost can be determined as C = $2000 + $100n.

The problem also states that the number of passengers n depends on the cost of the ticket n = 450 - 0.66T. So, let's put all of this together in the profit P equation:

P = R - C = (Tn) - (2000 + 100n) = Tn - 100n - 2000

I added parentheses to separate out the R and the C terms. Now lets take out the n from the two terms and substitute n = 450 - 0.66T into the equation.

P = n(T - 100) - 2000 = (450 - 0.66T)(T - 100) - 2000

We will need FOIL to distribute the two terms.

P = 450T - 45000 - 0.66T² + 66T - 2000

Simplifying this result, lets right the equation as a polynomial

P = -0.66T² + 516T - 47000

This is not the end. We need to determine the price T such that the profit P is maximized. This occurs when the derivative of the profit P equation is 0, so we need to find the derivative. The Power Rule is our friend here:

P' = -1.32T + 516

With the derivative at hand, lets solve for T when P' = 0

0 = -1.32T + 516

1.32T = 516

T = $390.91 approx.

However the question states to provide the answer to the nearest $5, so $390 is our profit-maximizing ticket price.

Let's check our work by plugging in values for T for the original profit equation P = R - C. I will use $390, $395 and $385. If our answer is correct, then $390 will result in a higher P than $395 or $385.

P(T) = -0.66T² + 516T - 47000

P(390) = -(0.66)(390²) + (516)(390) - 47000 = 53,854

P(395) = -(0.66)(395²) + (516)(395) - 47000 = 53,843.5

P(385) = -(0.66)(385²) + (516)(385) - 47000 = 53,831.5

As we can see, the price $390 yields the highest profit than its neighboring values of $395 or $385.