Let's break this down step by step.

(a) Finding the Cost Function $C(x)$

The cost function $C(x)$ consists of two components:

1. Fixed Costs: This is constant and equal to $54,000.
2. Variable Costs: This depends on the number of units produced, $x$. The variable cost per unit is given as: $\text{Variable Cost per Unit} = \frac{1}{3}x + 333$ Therefore, the total variable cost for $x$ units is: $\text{Total Variable Cost} = x \times \left(\frac{1}{3}x + 333\right)$

So, the cost function is the sum of the fixed and variable costs: $C(x) = 54000 + x\left(\frac{1}{3}x + 333\right)$ Simplify this expression: $C(x) = 54000 + \frac{1}{3}x^2 + 333x$

Thus, the simplified cost function is: $C(x) = \frac{1}{3}x^2 + 333x + 54000$

(b) Finding the Revenue Function $R(x)$

The revenue function $R(x)$ is the product of the number of units sold $x$ and the price per unit. The price per unit is given by: $\text{Price per Unit} = 2163 - \frac{2}{3}x$

So, the revenue function is: $R(x) = x \times \left(2163 - \frac{2}{3}x\right)$ Simplify the expression: $R(x) = 2163x - \frac{2}{3}x^2$

Thus, the simplified revenue function is: $R(x) = -\frac{2}{3}x^2 + 2163x$

(c) Finding the Profit Function $P(x)$

The profit function $P(x)$ is the difference between the revenue $R(x)$ and the cost $C(x)$: $P(x) = R(x) - C(x)$

Substitute the expressions for $R(x)$ and $C(x)$: $P(x) = \left(-\frac{2}{3}x^2 + 2163x\right) - \left(\frac{1}{3}x^2 + 333x + 54000\right)$ Distribute the negative sign and simplify: $P(x) = -\frac{2}{3}x^2 + 2163x - \frac{1}{3}x^2 - 333x - 54000$ Combine like terms: $P(x) = -\left(\frac{2}{3} + \frac{1}{3}\right)x^2 + (2163 - 333)x - 54000$ $P(x) = -x^2 + 1830x - 54000$

Thus, the simplified profit function is: $P(x) = -x^2 + 1830x - 54000$

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Let me know if you'd like further details on any part of the solution!

Here are 5 related questions:

1. How does the profit function change if fixed costs increase by $10,000?
2. What is the break-even point where profit $P(x) = 0$?
3. How does the variable cost per unit affect the cost function's shape?
4. What is the maximum revenue, and how can we find it?
5. How does changing the price per unit affect total revenue?

Tip: To maximize profit, you can use calculus to find the critical points of the profit function!