Market research indicates that consumers will buy x thousand units of a particular

kind of coffee maker when the unit price is p(x) = 0.27x + 51 dollars. The cost of

producing the x thousand units is C(x) = 2.23x

2 + 3.5x + 85 thousand dollars.

a. What are the revenue and profit functions, R(x) and P(x), for this production

process?

b. For what values of x is production of the coffee makers profitable?

Let's tackle this problem step by step.

### Part (a): Revenue and Profit Functions

1. **Revenue Function, \( R(x) \):**

   The revenue is calculated as the product of the number of units sold and the price per unit.

   \[
   R(x) = \text{{price per unit}} \times \text{{number of units sold}}
   \]

   Given the price function \( p(x) = 0.27x + 51 \) dollars and \( x \) is in thousands of units, the revenue function is:

   \[
   R(x) = x \times p(x) = x \times (0.27x + 51)
   \]

   Simplifying:

   \[
   R(x) = 0.27x^2 + 51x \text{ (in thousand dollars)}
   \]

2. **Profit Function, \( P(x) \):**

   The profit is the difference between revenue and cost.

   \[
   P(x) = R(x) - C(x)
   \]

   The cost function is given as \( C(x) = 2.23x^2 + 3.5x + 85 \) thousand dollars. So the profit function is:

   \[
   P(x) = (0.27x^2 + 51x) - (2.23x^2 + 3.5x + 85)
   \]

   Simplifying:

   \[
   P(x) = 0.27x^2 + 51x - 2.23x^2 - 3.5x - 85
   \]

   \[
   P(x) = (-1.96x^2 + 47.5x - 85) \text{ (in thousand dollars)}
   \]

### Part (b): Finding the Values of \( x \) for Profitability

To determine the values of \( x \) where production is profitable, we need to find when the profit function \( P(x) > 0 \).

\[
-1.96x^2 + 47.5x - 85 > 0
\]

This is a quadratic inequality. To solve it, first find the roots of the quadratic equation by setting \( P(x) = 0 \):

\[
-1.96x^2 + 47.5x - 85 = 0
\]

Let's solve this quadratic equation:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = -1.96 \), \( b = 47.5 \), and \( c = -85 \).

Now, I'll calculate the roots.The roots of the quadratic equation are approximately \( x_1 = 1.95 \) and \( x_2 = 22.29 \).

### Profitability Range

The profit function is a downward-opening parabola (since the coefficient of \( x^2 \) is negative), so the production will be profitable between these two roots:

\[
1.95 < x < 22.29
\]

Thus, production of the coffee makers is profitable when the number of units produced and sold (in thousands) is between approximately **1.95 thousand and 22.29 thousand units**.

---

Would you like more details on any part of the solution? Here are five related questions that could deepen your understanding:

1. How do we interpret the roots of the quadratic equation in real-world terms?
2. Can we calculate the maximum profit and at what value of \( x \) it occurs?
3. How do changes in the cost function affect the profitability range?
4. What would happen if the price function changed?
5. How can we graphically represent the profit function to visualize the profitability range?

**Tip:** Always consider the real-world context of the mathematical solutions, especially when dealing with quantities that represent physical objects or money.

Market research indicates that consumers will buy x thousand units of a particular kind of coffee maker when the unit price is p(x) = 0.27x + 51 dollars. The cost of producing the x thousand units is C(x) = 2.23x^2 + 3.5x + 85 thousand dollars. a. What are the revenue and profit functions, R(x) and P(x), for this production process? b. For what values of x is production of the coffee makers profitable?

Discover how to analyze the profitability of producing coffee makers using quadratic equations. This problem involves deriving revenue and profit functions, solving for profitability ranges, and applying the quadratic formula. Suitable for high school students in Grades 10-12.

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