The question asks how to determine the price that guarantees maximum revenue given the following conditions:

• Initial price: $55
• At $55, 402 units are sold.
• For every $1 increase in price, 6 fewer items are sold.

Step-by-Step Solution

Let:

• $x$ = number of dollars the price increases beyond $55
• $p(x) = 55 + x$ = new price after increasing by $x$
• $q(x) = 402 - 6x$ = number of items sold when price is increased by $x$

Revenue function:
$R(x) = \text{(Price)} \times \text{(Quantity Sold)} = (55 + x)(402 - 6x)$

Step 1: Expand the Revenue Function

$R(x) = 55(402 - 6x) + x(402 - 6x)$ $R(x) = 22110 - 330x + 402x - 6x^2$ $R(x) = -6x^2 + 72x + 22110$

Step 2: Find the Value of $x$ that Maximizes Revenue

This is a quadratic function in the form $R(x) = -6x^2 + 72x + 22110$.
Since the coefficient of $x^2$ is negative, the parabola opens downward, meaning the maximum revenue occurs at the vertex.

The formula for the $x$-coordinate of the vertex is:

$x = \frac{-b}{2a}$ where $a = -6$ and $b = 72$.

$x = \frac{-72}{2(-6)} = \frac{-72}{-12} = 6$

Step 3: Calculate the Optimal Price

The optimal price occurs when $x = 6$.
$\text{Optimal Price} = 55 + 6 = 61$

Final Answer

The price that guarantees maximum revenue is $61.

Would you like more details or clarifications?

Here are five related questions you might explore:

1. How do we know the revenue function opens downward?
2. What is the maximum revenue value for this problem?
3. How would the answer change if the number of items sold decreased by 10 instead of 6?
4. What happens if the price is increased by more than 6 dollars?
5. How can this concept be extended to solve profit maximization problems?

Tip: Quadratic functions are very useful for optimization problems, especially when maximizing or minimizing revenues and costs.