Answer:
A) The linear relation between price and demand is:
![d=-550x+2750](https://img.qammunity.org/2021/formulas/mathematics/college/v5k84w11zlj34467ftqbq6d3jdh4ts60xa.png)
The revenue R is:
![R=-550x^2+2750x](https://img.qammunity.org/2021/formulas/mathematics/college/qo41uogk2zd785td6ztpu57on8h98oal3e.png)
B) The profit functionP is:
![P=-550x^2+2750x-30](https://img.qammunity.org/2021/formulas/mathematics/college/jiq6s6049qi0yo8epfyzlspqsy6phu6q6h.png)
C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.
Explanation:
We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.
We have two points for this linear relationship:
- At price x=3, the demand is d=1100.
- At price x=2.5, the demand is d=1375.
We will model the relation:
![d=mx+b](https://img.qammunity.org/2021/formulas/mathematics/college/79v5o21s45fcker7otjz2laf01swaj0y7z.png)
We can calculate the slope m as:
![m=(\Delta d)/(\Delta x)=(d_2-d_1)/(x_2-x_1)=(1375-1100)/(2.5-3)\\\\\\m=(275)/(-0.5)=-550](https://img.qammunity.org/2021/formulas/mathematics/college/utyk3qx2q3nftne4675qpkq3csiaru2pc8.png)
Then, replacing one point in the linear equation, we can calculate the intercept b:
![d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750](https://img.qammunity.org/2021/formulas/mathematics/college/xdydz47okdbe2huzp7tlr45nhu7wo25zr8.png)
Then, the linear relation between demand and price is:
![d=-550x+2750](https://img.qammunity.org/2021/formulas/mathematics/college/v5k84w11zlj34467ftqbq6d3jdh4ts60xa.png)
The revenue R can be expressed as the multiplication of the price and the demand:
![R=x\cdot d=x(-550x+2750)=-550x^2+2750x](https://img.qammunity.org/2021/formulas/mathematics/college/8qrpuacqpuws8wnbtq9s5pg9s6nb21eo4c.png)
If we have a fixed cost of $30 per month, the profit P is:
![P=R-FC=-550x^2+2750x-30](https://img.qammunity.org/2021/formulas/mathematics/college/mhxnmxqks6l8ahzln3qf8k9hreoiw6h44y.png)
We can maximize the profit by deriving the profit function and making it equal to zero.
![(dP)/(dx)=0\\\\\\(dP)/(dx)=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=(2750)/(1100)=2.5](https://img.qammunity.org/2021/formulas/mathematics/college/brxqnxtrewjiwyeeytk6bfvtg7t03vjom4.png)
This corresponds to a profit of:
![P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5](https://img.qammunity.org/2021/formulas/mathematics/college/k8mz3fb1ikguko3dw07yhdfs48icab7oha.png)