Answer:
A) The linear relation between price and demand is:
![d=-550x+2750](https://tex.z-dn.net/?f=d%3D-550x%2B2750)
The revenue R is:
![R=-550x^2+2750x](https://tex.z-dn.net/?f=R%3D-550x%5E2%2B2750x)
B) The profit functionP is:
![P=-550x^2+2750x-30](https://tex.z-dn.net/?f=P%3D-550x%5E2%2B2750x-30)
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.
Step-by-step 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://tex.z-dn.net/?f=d%3Dmx%2Bb)
We can calculate the slope m as:
![m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7B%5CDelta%20d%7D%7B%5CDelta%20x%7D%3D%5Cdfrac%7Bd_2-d_1%7D%7Bx_2-x_1%7D%3D%5Cdfrac%7B1375-1100%7D%7B2.5-3%7D%5C%5C%5C%5C%5C%5Cm%3D%5Cdfrac%7B275%7D%7B-0.5%7D%3D-550)
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://tex.z-dn.net/?f=d_1%3Dmx_1%2Bb%5C%5C%5C%5C1100%3D%28-550%29%5Ccdot%203%2Bb%5C%5C%5C%5C1100%3D-1650%2Bb%5C%5C%5C%5Cb%3D1100%2B1650%3D2750)
Then, the linear relation between demand and price is:
![d=-550x+2750](https://tex.z-dn.net/?f=d%3D-550x%2B2750)
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://tex.z-dn.net/?f=R%3Dx%5Ccdot%20d%3Dx%28-550x%2B2750%29%3D-550x%5E2%2B2750x)
If we have a fixed cost of $30 per month, the profit P is:
![P=R-FC=-550x^2+2750x-30](https://tex.z-dn.net/?f=P%3DR-FC%3D-550x%5E2%2B2750x-30)
We can maximize the profit by deriving the profit function and making it equal to zero.
![\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5](https://tex.z-dn.net/?f=%5Cdfrac%7BdP%7D%7Bdx%7D%3D0%5C%5C%5C%5C%5C%5C%5Cdfrac%7BdP%7D%7Bdx%7D%3D-550%282x%29%2B2750%281%29%3D0%5C%5C%5C%5C%5C%5C-1100x%2B2750%3D0%5C%5C%5C%5Cx%3D%5Cdfrac%7B2750%7D%7B1100%7D%3D2.5)
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://tex.z-dn.net/?f=P%282.5%29%3D-550%282.5%29%5E2%2B2750%282.5%29-30%5C%5C%5C%5CP%282.5%29%3D-550%5Ccdot%206.25%2B6875-30%5C%5C%5C%5CP%282.5%29%3D-3437.5%2B6875-30%5C%5C%5C%5CP%282.5%29%3D3407.5)