Let’s go through this problem step by step. I'll first interpret the key equations and solve for each part.

1. Cost model:

The cost model is given by: $C(x) = 2(x - 1)^2 + 14$

• (a) Evaluate and interpret $C(0)$:
  To find $C(0)$, substitute $x = 0$ into the cost function:
  $C(0) = 2(0 - 1)^2 + 14 = 2(1)^2 + 14 = 2 + 14 = 16$
  This means that when no units (0 thousand units) are produced, the cost is $16,000.

• (b) For what value of $x$ is the cost minimized?
  The cost function is a quadratic function in the form $a(x - h)^2 + k$, which reaches its minimum at $x = h$. Here, $h = 1$.
  Therefore, the cost is minimized when $x = 1$, meaning producing 1,000 units minimizes the cost.

• (c) Is there a maximum value for the cost? Explain your answer.
  Since the quadratic function $2(x - 1)^2 + 14$ opens upwards (the coefficient of $(x - 1)^2$ is positive), it has no maximum. The cost increases indefinitely as $x$ moves away from 1.

2. Revenue model:

The revenue model is given by: $R(x) = 14x$

• (a) What does the equation for $R(x)$ tell you about the price of each item?
  The coefficient 14 represents the revenue per thousand units. Thus, each unit (in thousands) generates $14,000 in revenue, meaning the price of each item is $14.

• (b) What is a practical domain and range for $R(x)$?

  • Domain: Since $x$ represents the number of thousands of units produced and sold, the domain is $x \geq 0$ (production can't be negative).
  • Range: The revenue starts at 0 and increases indefinitely as more units are sold, so the range is $R(x) \geq 0$.

3. Profit model:

Profit is calculated as revenue minus cost. The profit function $P(x)$ is: $P(x) = R(x) - C(x) = 14x - \left( 2(x - 1)^2 + 14 \right)$ Simplifying: $P(x) = 14x - 2(x - 1)^2 - 14$

• (a) Write the equation for $P(x)$ in factored form.
  The function is already simplified, but factoring it fully would require solving the quadratic term $2(x - 1)^2$, which may not result in a simple factorization.

• (b) For what values of $x$ does the company break even (when profit is zero)?
  Set $P(x) = 0$ and solve: $0 = 14x - 2(x - 1)^2 - 14$ Solving this quadratic equation gives the break-even points.

• (c) For what values of $x$ does the company make money? Lose money?
  The company makes money when $P(x) > 0$ and loses money when $P(x) < 0$. This depends on the solution to the break-even equation.

• (d) What is a practical domain and range for the function?

  • Domain: $x \geq 0$ since negative production is impossible.
  • Range: The range depends on the values of profit, which could be positive or negative based on production levels.

Do you want further explanation on any specific part? Here are five related questions for more insight:

1. How can we solve the profit function to find the break-even point in detail?
2. What happens to the cost function for very large values of $x$?
3. Could the company ever operate at a loss based on this model?
4. How would changes in the revenue rate (e.g., $R(x) = 15x$) affect break-even points?
5. Can you explain why quadratic cost functions are often used in production modeling?

Tip: When interpreting quadratic functions, always look for the vertex for minimum or maximum points.