To solve this problem, we need to find the price that will maximize the revenue for The Better Baby Buggy Co.

Step 1: Express the Revenue Function

Revenue $R$ is the product of price $p$ and the quantity sold $q$. We are given the demand equation: $q = -3p + 294$ Thus, revenue $R$ is: $R = p \cdot q$ Substitute the expression for $q$ into the revenue equation: $R = p \cdot (-3p + 294)$ $R = -3p^2 + 294p$

Step 2: Maximize the Revenue Function

This is a quadratic equation of the form $R(p) = -3p^2 + 294p$, which is a downward-facing parabola (since the coefficient of $p^2$ is negative). The maximum value of a parabola occurs at its vertex. The formula to find the vertex of a parabola $ax^2 + bx + c$ is: $p = \frac{-b}{2a}$ Here, $a = -3$ and $b = 294$. Substituting these values into the vertex formula: $p = \frac{-294}{2(-3)} = \frac{294}{6} = 49$

Thus, the price that maximizes revenue is $49.

Step 3: Calculate the Maximum Revenue

Now that we know the price $p = 49$, substitute it back into the revenue equation to find the maximum revenue: $R = -3(49)^2 + 294(49)$ First, calculate $49^2$ and then substitute: $49^2 = 2401$ $R = -3(2401) + 294(49)$ $R = -7203 + 14406$ $R = 7203$

Thus, the maximum monthly revenue is $7203.

Final Answers:

• The company should sell the buggies at $49 per buggy to maximize revenue.
• The largest monthly revenue is $7203.

Would you like more details or have any questions?

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Here are some related questions:

1. How would the revenue change if the price is set higher or lower than $49?
2. What happens to the demand for the Turbo buggies if the price is increased by $10?
3. How would an increase in the coefficient of $p$ in the demand equation affect revenue?
4. If fixed costs are introduced, how does that change the maximum profit calculation?
5. Can you express the relationship between price and revenue graphically?

Tip: The vertex of a quadratic function provides the maximum or minimum point, depending on the direction of the parabola.