A coffee shop currently sells 470 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, NN , is linearly related to the sale price, pp  (in dollars). Find an equation for *NN*  as a function of *pp* .

*N(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 *pp*  as the sales price, use your equation from above to write an equation for the revenue, *RR*  as a function of *pp* .

*R(p)R(p)*  =   

c) The store wants to maximize their revenue (make as much money as possible). Find the value of *pp*  that will maximize the revenue (round to the nearest cent).

*p* =   
which will give a maximum revenue of $  

Let's work through the problem step by step.

### Part (a): Find the equation for $ N(p) $, the number of lattes sold per day as a function of the price $ p $.

Given:
- Initially, the coffee shop sells 470 lattes at a price of $2.75.
- When the price is increased by $0.25 (making it $3.00), the number of lattes sold decreases by 40, so they sell 430 lattes at $3.00.

Assume that the relationship between the number of lattes sold $ N $ and the price $ p $ is linear. The equation of a line can be written in the form:
\[
N(p) = mp + b
\]
where $ m $ is the slope and $ b $ is the y-intercept.

**Step 1: Find the slope $ m $.**

The slope $ m $ is calculated as the change in the number of lattes sold divided by the change in price:
\[
m = \frac{\Delta N}{\Delta p} = \frac{430 - 470}{3.00 - 2.75} = \frac{-40}{0.25} = -160
\]

So, the slope $ m = -160 $.

**Step 2: Find the y-intercept $ b $.**

Substitute one of the points (for example, $ (p, N) = (2.75, 470) $) into the linear equation to solve for $ b $:
\[
470 = -160(2.75) + b
\]
\[
470 = -440 + b
\]
\[
b = 910
\]

Thus, the equation for $ N(p) $ is:
\[
N(p) = -160p + 910
\]

### Part (b): Write the equation for revenue $ R(p) $ as a function of the price $ p $.

Revenue $ R $ is the product of the number of lattes sold and the price per latte:
\[
R(p) = p \times N(p)
\]

Substitute the equation for $ N(p) $ from part (a):
\[
R(p) = p \times (-160p + 910)
\]

Distribute $ p $:
\[
R(p) = -160p^2 + 910p
\]

### Part (c): Find the price $ p $ that maximizes the revenue $ R(p) $.

To find the price that maximizes revenue, we need to find the vertex of the parabola described by $ R(p) = -160p^2 + 910p $. The price $ p $ at the vertex of a parabola can be found using the formula:
\[
p = -\frac{b}{2a}
\]
where $ R(p) = ap^2 + bp + c $.

Here, $ a = -160 $ and $ b = 910 $. Substituting these values:
\[
p = -\frac{910}{2(-160)} = \frac{910}{320} = 2.84375
\]

Rounding to the nearest cent:
\[
p \approx 2.84 \text{ dollars}
\]

So, the price $ p $ that maximizes revenue is **$2.84**.

**Step 3: Find the maximum revenue.**

Substitute $ p = 2.84 $ back into the revenue equation $ R(p) $:
\[
R(2.84) = -160(2.84)^2 + 910(2.84)
\]

Calculate this:
\[
R(2.84) = -160(8.0656) + 910(2.84)
\]
\[
R(2.84) = -1290.496 + 2584.4 = 1293.904
\]

So, the maximum revenue is approximately **$1,293.90**.

### Summary:
- **(a)** The equation for the number of lattes sold is $ N(p) = -160p + 910 $.
- **(b)** The revenue function is $ R(p) = -160p^2 + 910p $.
- **(c)** The price that maximizes revenue is $2.84, resulting in a maximum revenue of approximately $1,293.90.

Do you want more details on any part of this solution? Here are five related questions:

1. How does changing the slope in the linear equation $ N(p) $ affect the revenue?
2. What would happen to the revenue if the price were set lower than $2.84?
3. How would you interpret the y-intercept $ b = 910 $ in the context of the problem?
4. Can you use calculus (taking the derivative) to confirm the maximum revenue?
5. How does the quadratic nature of the revenue function influence pricing strategy?

**Tip:** When maximizing or minimizing a quadratic function, the vertex formula $ p = -\frac{b}{2a} $ is a quick way to find the optimal point without needing to complete the square or use calculus.

Explore how a coffee shop maximizes revenue by analyzing the impact of a price increase on latte sales. Discover the linear relationship between price and sales, and the quadratic equation used to determine optimal pricing for maximum revenue.

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