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.