So, to maximize monthly revenue, the company should charge $9 per month, and the maximum revenue will be $121.5 million.
To maximize monthly revenue, we need to find the price at which the revenue is maximized. Revenue is calculated by multiplying the price per subscriber by the number of subscribers.
Let P be the price increase in dollars, and Q be the number of subscribers.
The original price is $8, and the original number of subscribers is 15 million.
So, the original revenue
is given by:
![\[ R_0 = P_0 * Q_0 = 8 * 15 \, \text{million} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/nx4guctmbv893hb80wkyvirfkfz89o6wys.png)
Now, for each $1 increase in price, the company loses 1.5 million subscribers. So, if the price increases by P dollars, the number of subscribers Q will be
million.
The revenue R with the price increase is given by:
![\[ R = (8 + P) * (15 - 1.5P) \, \text{million} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/f5q24vff5fffh1449aunqfo72kwheojzda.png)
Now, we want to find the value of P that maximizes R. To do this, we can find the critical points by taking the derivative of R with respect to P and setting it equal to zero:
![\[ (dR)/(dP) = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/q5c7c0sjxlzu3g3jdeebarqculp094mdx1.png)
![\[ (d)/(dP) [(8 + P) * (15 - 1.5P)] = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/c79rerm2jbz3engbn14lbz7bvgcdyh4loi.png)
![\[ (15 - 1.5P) - (8 + P) * 1.5 = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ix66txi4ozpvpblo84ecroj9732h7dddg7.png)
Sure, let's solve for P in the equation:
![\[ (15 - 1.5P) - (8 + P) * 1.5 = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ix66txi4ozpvpblo84ecroj9732h7dddg7.png)
First, distribute
on the right side:
![\[ 15 - 1.5P - 12 - 1.5P = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/ulf2j3rwlor1ktv8wzff8j0w4jwef60dh8.png)
Combine like terms:
![\[ 3 - 3P = 0 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/9epqnr1azsypkp1gkbkw5y0vi0jdbqlrug.png)
Now, isolate
:
![\[ -3P = -3 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/wmy9kneoqugrbwa7m7yi623ssv1gel52dn.png)
Divide both sides by
:
![\[ P = 1 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/6xpplwzonqg7sq8su7snln2y6qlcjnp11g.png)
So,
is the solution. Now, we need to determine the corresponding price and calculate the revenue.
The original price is $8, and for each $1 increase, the new price is
dollars.
Now, calculate the revenue
:
![\[ R = (8 + P) * (15 - 1.5P) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/el17bkqmrhvxmmfausr4hjm96pylxna8zt.png)
![\[ R = 9 * (15 - 1.5 * 1) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/6u2xm72cpo0p82zpzi929r4d1va2pr5yz5.png)
![\[ R = 9 * 13.5 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/iuwjtswm9lqu7m76827vxshoa7szxkidqe.png)
![\[ R = 121.5 \, \text{million dollars} \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/dmt1wy2qmoir2g0rrx1n851ee3y8mqxect.png)