# Calculus optimization quick question

A hotel fills only $120$ rooms when the price is $\$150$per night for a room. When the price is decreased by$\$10$, it fills $16$ additional rooms.

Find the price for maximum revenue.

Ok frankly speaking I can do the question, but I just don't understand this little part :-

Suppose $p(x)$ is the price function,

$p(x) = 150 - \frac{10}{16}(x-120)$ , where $x$ = increased total number of rooms

I understand that $(x-120)$ gives the increase in number of rooms filled, and $\frac{10}{16}$ is the price decrease per each room, but I don't know what does $150 - \frac{10}{16}(x-120)$ tell me? Someone lighten me up?

• I'll warn you for future reference that the "dollar sign" is a special character in MathJax, so you cannot use it directly in your problem statements. – colormegone Jun 12 '14 at 6:01

I'm not sure what you want answered, so I will give you more questions that will guide you towards whatever answer it is you want to find.

You're interested in a price function, $p(x)$, that tells you the price per room that results in $x$ rooms being filled.

$p(x)$ is the inverse function of the room function, $r(x)$, that tells you how many rooms will be filled as a result of setting the price per room to $x$.

Problem:

You're told that when the price is 150 per room, 120 rooms end up being filled. You're also told that decreasing the price per room by 10 increases the number of rooms being filled by 16; in fact, a price decrease by $a\cdot10$ for any not-necessarily-whole scale factor $a$ results in $a\cdot16$ additional rooms being filled. For the purposes of this problem, we'll allow a non-whole number of rooms to be filled.

Part a. Find $r(x)$.

If $x$ is the price per room, then the number of rooms that are filled is given by

$$r(x) = 120 - (x-150) \cdot 16/10$$

The higher the price $x$ is from 150, the lower that $r(x)$ becomes. When $x$ is exactly 150, $r(x)$ is exactly 120.

Part b. Find $p(x)$.

To determine $p(x)$, we swap the input and output of $r(x)$:

$$x = 120 - (p(x)-150) \cdot 16/10 \\ x + (p(x)-150) \cdot 16/10 = 120 \\ p(x) = (120 - x) \cdot 10/16 + 150$$

$p(x)$ now tells you the price per room that results in $x$ rooms being filled. The higher the $x$ you want, the lower that $p(x)$ reports to you; this reflects the fact that to attract more visitors, you would need to set the price lower.

Part c. The hotel's income is given by

$$\text{price per room} \times \text{number of rooms filled} = p \cdot r.$$

Use $r(x)$ to create the function $I_1(x)$ that finds the hotel's income for a given price per room $x$, and use this function to find the price per room $p$ that would optimize the hotel's income. Then find $I_1(p)$ to find the hotel's optimized income.

Hint: $I_1(x)$ is at its maximum when its derivative with respect to $x$ is 0.

Part d. Use $p(x)$ rather than $r(x)$ to create the function $I_2(x)$ that gives the hotel's income for a given number of rooms $x$, and use this function to find the number of rooms $r$ that optimizes the hotel's revenue. Then find $I_2(r)$ to find the hotel's optimized income, and verify that the result matches that found in part c (i.e. $I_2(r) = I_1(p)$).

Hint: $I_2(x)$ is a different function from $I_1(x)$ (since $x$ takes on different meanings between the two functions), but it is also optimized when its derivative with respect to $x$ is 0.