A charter flight charges a fare of $300 per person plus $10 per person for each unsold seat on the plane. The plane holds 100 passengers. The total revenue received can be modelled by R(x)=-10x squared 700x 30,000, where x represents the number of unsold seats. The maximum revenue is $__ ,which occurs when __ seats are left unsold. How do I find the maximum revenue and the number of seats left unsold?

You can do it in two different ways. In both ways, you are finding the maximum point of the curve:

y = -10x^2 + 700x + 30000

1. Completing the square:

R(x) = -10x^2 + 700x + 30000

= -10(x^2 - 70x) + 30000

= -10[(x-35)^2 - 1225] + 30000

= -10(x-35)^2 + 12250+ 30000

= -10(x-35)^2 + 42250

From this, you know that the optimum occurs at coordinates (35, 42250)

Hence, it occurs when the number of empty seats is 35 and the revenue is $42250

2. Finding the derivative.

R(x) = -10x^2 + 700x + 30000

R'(x) = -20x + 700

At maximum, R'(x) = 0, so:

0 = -20x + 700

20x = 700

x = 35

Using this value of x in the formula,

R(35) = -10(35)^2 + 700(35) + 30000 = 42250

There you are :)

Given an equation such as


R = -10x^2 +700x+30000


you find the function's minimum and/or maximum values by determining where the slope of the function is equal to zero. So for this problem:


\frac {dR} {dx} = 700 - 20x = 0


Solve for x. Then to see whether this point corresponds to a minimum or a maximum take the second derivative of the function and evaluate it at this value of x. If the second derivative is positive you have a local minimum; if it's negative you have a local maximum. Finally, sub that value of x back into your original equation to get the corresponding value of R.

Please post back with what you get for answers.

Thank you so much!

I got 35 and 42250