A coffee shop currently sells 420 lattes a day at $2.50 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 $  

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Box 1: Enter your answer as an expression. Example: 3x^2+1, x/5, (a+b)/c
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Box 3: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 2^3, 5+4)
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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