A coffee shop currently sells 480 lattes a day at $\$ 2.75$ each. They recently tried raising the by price by $\$ 0.25$ a latte, and found that they sold 40 less lattes a day. a) Assume that the number of lattes they sell in a day, $N$, is linearly related to the sale price, $p$ (in dollars). Find an equation for $N$ as a function of $p$. \[ N(p)= \] b) Revenue (the amount of money the store brings in before costs) can be found by multiplying the cost per cup times the number of cups sold. Again using $p$ as the sales price, use your equation from above to write an equation for the revenue, $R$ as a function of $p$. \[ R(p)= \] c) The store wants to maximize their revenue (make as much money as possible). Find the value of $p$ that will maximize the revenue (round to the nearest cent). \[ p= \] which will give a maximum revenue of $\$$

Step 1 ：We are given two points on the line that describes the relationship between the number of lattes sold and the price. The points are (2.75, 480) and (3, 440). We can use these points to find the slope of the line, which is the change in the number of lattes sold divided by the change in price. Once we have the slope, we can use one of the points to find the y-intercept of the line, which is the number of lattes that would be sold if the price were zero. This will give us the equation for N(p).

Step 2 ：The slope of the line is -160 and the y-intercept is 920. This means that for every $1 increase in price, they sell 160 less lattes. If the price were $0, they would sell 920 lattes. Therefore, the equation for N(p) is \(N(p) = -160p + 920\).

Step 3 ：Next, we need to find the equation for the revenue, R(p). The revenue is the price per latte times the number of lattes sold, so we can substitute the equation for N(p) into this formula to get \(R(p) = p * N(p) = p * (-160p + 920)\).

Step 4 ：Finally, to maximize the revenue, we need to find the value of p that gives the maximum value of R(p). This can be found by taking the derivative of R(p) with respect to p, setting it equal to zero, and solving for p. This will give us the price that maximizes revenue.

Step 5 ：The price that will maximize the revenue is \(\boxed{\$2.88}\), which will give a maximum revenue of \(\boxed{\$1322.50}\).