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


        They lose 30 customers for each raising the price of $0.25.   It means that the rate of losing customers 
        is  30*4 = 120 per dollar raising. So

            N(p) = 410 + 120*(2.75 - p),      (1)

        where "p" is the new price.   Part (a) is completed.



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.


        Revenue is the product of the price by the number of customers

            R(p)= p*N(p) = p*(410 + 120*(2.75 - p)),    (2)

        according to (1).


        So, the equation is obtained and part (b) is completed.



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).

        
        The revenue, as I showed you in part (b) (formula (2)) is this quadratic function

            R(p) = p*(410 + 120*(2.75 - p)) = -120*p^2 + 740p.    (3)


        This quadratic function has a negative leading coefficient at p^2; so the parabola is open downward and has a maximum.


        For any such parabola, its maximum is achieved at  p = ,

            where "a" is the coefficient at the quadratic term and "b" is the coefficient at the linear term.


        In your case,  a = -120,  b = 740,  therefore the optimum value of p is  

             =  =  =  = 3.08 dollars (rounded to two decimals).   ANSWER


        The optimum price is 3.08 dollars per latte, and the maximum revenue is then  -120*3.08^2 + 740*3.08 = 1140.83 dollars.

        Compare it with the current revenue of  410*2.75 = 1127.50 dollars.



    See the plot below.


    


                Plot y = x*(410 + 120*(2.75 - x))