Application with differential calculus

Question

An airline will fill $100$ seats of its aircraft at a fare of $\$200.$ For every $ \$5$ increase in the fare, the plane loses two passengers. For every decrease of $\$5$, the company gains two passengers. What price maximizes revenue?

Answer 1

$$\text{Revenue} = (200+5k)(100-2 k).$$

To maximize revenue, find its derivative with respect to $k$. To know you are at a maximum, rather than at a minimum, look at the second derivative. Alternatively, graph the function.

By the way, one subtlety is that the number of passengers is integer. You might want to think about why this method works, nevertheless.

UPDATE

As @AndrewChin wrote in a comment, the most straightforward way to look at this is to understand that this is a quadratic function (a parabola if you expand the support to the reals). Writing the revenue as:

$$\text{Revenue} = 10(2025-(5-k)^2),$$ it should be clear what $k$ maximizes this.