To solve this problem, we'll derive the revenue function $R(x)$ and the profit function $P(x)$ using the given information about price and cost.

1. Revenue Function $R(x)$:

Revenue is the product of the price per unit and the number of units sold.

The price function is given as: $p(x) = -0.27x + 51$ Here, $x$ represents the number of thousand units sold, and $p(x)$ is the price per unit in dollars.

The revenue $R(x)$ is the total income from selling $x$ thousand units, and it's calculated by multiplying the price per unit by the number of units sold: $R(x) = x \cdot p(x)$

Substitute $p(x)$ into this equation: $R(x) = x \cdot (-0.27x + 51)$

Simplify: $R(x) = -0.27x^2 + 51x$

Thus, the revenue function is: $R(x) = -0.27x^2 + 51x \quad \text{(in thousand dollars)}$

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2. Profit Function $P(x)$:

Profit is the difference between revenue and cost.

The cost function is given as: $C(x) = 2.23x^2 + 3.5x + 85$ where $C(x)$ represents the cost in thousand dollars for producing $x$ thousand units.

The profit function $P(x)$ is calculated by subtracting the cost from the revenue: $P(x) = R(x) - C(x)$

Substitute $R(x)$ and $C(x)$ into this equation: $P(x) = (-0.27x^2 + 51x) - (2.23x^2 + 3.5x + 85)$

Simplify: $P(x) = -0.27x^2 + 51x - 2.23x^2 - 3.5x - 85$ $P(x) = (-0.27x^2 - 2.23x^2) + (51x - 3.5x) - 85$ $P(x) = -2.5x^2 + 47.5x - 85$

Thus, the profit function is: $P(x) = -2.5x^2 + 47.5x - 85 \quad \text{(in thousand dollars)}$

Summary:

• The revenue function is $R(x) = -0.27x^2 + 51x$.
• The profit function is $P(x) = -2.5x^2 + 47.5x - 85$.

Would you like further details or have any questions about the process?

Here are 5 related questions for you:

1. How many units should be produced to maximize revenue?
2. At what production level does the profit equal zero?
3. How would the revenue function change if the price per unit decreased?
4. Can the cost function be optimized to increase profit?
5. What is the maximum profit possible for this production process?

Tip: To maximize profit or revenue, you can use calculus techniques such as finding the derivative of the function and setting it equal to zero.