Try another version of this question

A coffee shop currently sells 460 lattes a day at $2.75 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 30 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 $

Get help:

Box 1: Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c

Be sure your variables match those in the question

Box 2: Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c

Be sure your variables match those in the question

Box 3: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)

Enter DNE for Does Not Exist, oo for Infinity

Box 4: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)

Enter DNE for Does Not Exist, oo for Infinity