The market research department of the Better Baby Buggy Co. predicts that the demand equation for its buggies is given by
q = −3.5p + 1050
where q is the number of buggies the company can sell in a month if the price is $p per buggy. At what price (in dollars) should it sell the buggies to get the largest revenue?
p = $
What is the largest monthly revenue (in dollars)?

To solve this problem, we need to determine the price $ p $ that will maximize the monthly revenue. The revenue $ R $ is calculated as:

\[
R(p) = p \times q
\]
where $ q $ is the number of buggies sold at price $ p $, and the demand equation is:

\[
q = -3.5p + 1050
\]

### Step 1: Express the revenue function
Substitute the expression for $ q $ into the revenue equation:

\[
R(p) = p \times (-3.5p + 1050)
\]

This simplifies to:

\[
R(p) = -3.5p^2 + 1050p
\]

### Step 2: Find the price that maximizes revenue
The revenue function $ R(p) $ is a quadratic equation, and since the coefficient of $ p^2 $ is negative, the parabola opens downwards. This means the maximum revenue occurs at the vertex of the parabola. The formula for the vertex of a quadratic function $ ax^2 + bx + c $ is given by:

\[
p = -\frac{b}{2a}
\]

For our equation $ R(p) = -3.5p^2 + 1050p $, $ a = -3.5 $ and $ b = 1050 $. Substituting these values:

\[
p = -\frac{1050}{2(-3.5)} = \frac{1050}{7} = 150
\]

So, the company should set the price at **$150** to maximize revenue.

### Step 3: Calculate the maximum revenue
Now substitute $ p = 150 $ into the revenue function $ R(p) $:

\[
R(150) = -3.5(150)^2 + 1050(150)
\]

First, calculate $ 150^2 = 22500 $:

\[
R(150) = -3.5(22500) + 1050(150)
\]
\[
R(150) = -78750 + 157500 = 78750
\]

So, the maximum monthly revenue is **$78,750**.

### Final Answer:
- Price: **$150**
- Maximum monthly revenue: **$78,750**

Let me know if you need more details or have any questions!

Here are some related questions:
1. How does changing the price affect the demand based on the equation?
2. What happens to the revenue if the price is set lower than $150?
3. What are the key components that define a quadratic revenue function?
4. Can the company increase sales beyond 1050 buggies per month?
5. What happens to the revenue when the price is very high, say $500?

**Tip**: In quadratic optimization problems, identifying the vertex of the parabola is crucial to finding the maximum or minimum value.

The market research department of the Better Baby Buggy Co. predicts that the demand equation for its buggies is given by q = −3.5p + 1050 where q is the number of buggies the company can sell in a month if the price is $p per buggy. At what price (in dollars) should it sell the buggies to get the largest revenue? p = $ What is the largest monthly revenue (in dollars)?

This detailed solution explains how to determine the optimal price for maximizing revenue for Better Baby Buggy Co. using the demand equation q = -3.5p + 1050. Learn how to apply the quadratic function R(p) = -3.5p^2 + 1050p and find the maximum revenue of $78,750 when the price is set at $150.

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