A small family firm manufacturing hand-made ceramic vases knows that it has the cost 
function 
C(x)
and the demand function 
D(x)
given by 
C(x) = 2000 + 20x and D(x) = 200 − x
where 
x
is the number of vases 
manufactured and sold.
Notes : Revenue, 
R(x) = xD(x)
and Profit, 
P(x) = R(x) −C(x).
Find 
(a) the revenue function and the level of production at which revenue is maximized,
(b) the level of production at which profit is maximized and the value of the 
maximum profit,
(c) the selling price in order to obtain maximum profit,
(d) the maximum profit if each ceramic vase is subject to a sales tax of RM20.

To solve this problem step by step, we will use the given functions and apply some basic calculus to optimize revenue and profit. Let's break it down:

### Given:
- Cost function: \( C(x) = 2000 + 20x \)
- Demand function: \( D(x) = 200 - x \)
- Revenue function: \( R(x) = xD(x) \)
- Profit function: \( P(x) = R(x) - C(x) \)

Where \( x \) represents the number of vases manufactured and sold.

### Part (a): Find the revenue function and the level of production at which revenue is maximized.

1. **Revenue Function**:
   The revenue \( R(x) \) is given by the formula:
   \[
   R(x) = xD(x) = x(200 - x)
   \]
   Simplifying:
   \[
   R(x) = 200x - x^2
   \]
   So, the revenue function is:
   \[
   R(x) = 200x - x^2
   \]

2. **Maximizing Revenue**:
   To find the level of production that maximizes revenue, we need to take the derivative of \( R(x) \) with respect to \( x \), and set it equal to zero.
   \[
   \frac{dR(x)}{dx} = 200 - 2x
   \]
   Set the derivative equal to zero:
   \[
   200 - 2x = 0
   \]
   Solving for \( x \):
   \[
   x = 100
   \]
   To confirm that this is a maximum, we check the second derivative of \( R(x) \):
   \[
   \frac{d^2R(x)}{dx^2} = -2
   \]
   Since the second derivative is negative, this confirms that \( x = 100 \) is the level of production that maximizes revenue.

Thus, the **revenue function is \( R(x) = 200x - x^2 \), and revenue is maximized when 100 vases are produced**.

### Part (b): The level of production at which profit is maximized and the value of the maximum profit.

1. **Profit Function**:
   The profit function is:
   \[
   P(x) = R(x) - C(x)
   \]
   Substituting the revenue and cost functions:
   \[
   P(x) = (200x - x^2) - (2000 + 20x)
   \]
   Simplifying:
   \[
   P(x) = 200x - x^2 - 2000 - 20x
   \]
   \[
   P(x) = 180x - x^2 - 2000
   \]
   So, the profit function is:
   \[
   P(x) = -x^2 + 180x - 2000
   \]

2. **Maximizing Profit**:
   To find the level of production that maximizes profit, take the derivative of \( P(x) \) with respect to \( x \) and set it equal to zero:
   \[
   \frac{dP(x)}{dx} = 180 - 2x
   \]
   Set the derivative equal to zero:
   \[
   180 - 2x = 0
   \]
   Solving for \( x \):
   \[
   x = 90
   \]
   To confirm that this is a maximum, check the second derivative of \( P(x) \):
   \[
   \frac{d^2P(x)}{dx^2} = -2
   \]
   Since the second derivative is negative, \( x = 90 \) is indeed the level of production that maximizes profit.

3. **Maximum Profit**:
   Substitute \( x = 90 \) into the profit function to find the maximum profit:
   \[
   P(90) = -(90)^2 + 180(90) - 2000
   \]
   \[
   P(90) = -8100 + 16200 - 2000 = 6100
   \]
   Thus, the **maximum profit is RM6100** when 90 vases are produced.

### Part (c): The selling price in order to obtain maximum profit.

The selling price is given by the demand function \( D(x) \), which represents the price per vase. At maximum profit, \( x = 90 \), so the selling price is:
\[
D(90) = 200 - 90 = 110
\]
Thus, the **selling price is RM110** per vase to achieve maximum profit.

### Part (d): The maximum profit if each ceramic vase is subject to a sales tax of RM20.

If each vase is subject to a sales tax of RM20, the firm's revenue per vase decreases by RM20. So, the new revenue function becomes:
\[
R_{\text{new}}(x) = x(D(x) - 20) = x(200 - x - 20) = x(180 - x)
\]
Simplifying:
\[
R_{\text{new}}(x) = 180x - x^2
\]

1. **New Profit Function**:
   The new profit function is:
   \[
   P_{\text{new}}(x) = R_{\text{new}}(x) - C(x)
   \]
   Substituting the new revenue and cost functions:
   \[
   P_{\text{new}}(x) = (180x - x^2) - (2000 + 20x)
   \]
   Simplifying:
   \[
   P_{\text{new}}(x) = 160x - x^2 - 2000
   \]

2. **Maximizing New Profit**:
   Take the derivative of \( P_{\text{new}}(x) \) with respect to \( x \):
   \[
   \frac{dP_{\text{new}}(x)}{dx} = 160 - 2x
   \]
   Set the derivative equal to zero:
   \[
   160 - 2x = 0
   \]
   Solving for \( x \):
   \[
   x = 80
   \]

3. **Maximum New Profit**:
   Substitute \( x = 80 \) into the new profit function:
   \[
   P_{\text{new}}(80) = -(80)^2 + 160(80) - 2000
   \]
   \[
   P_{\text{new}}(80) = -6400 + 12800 - 2000 = 4400
   \]
   Thus, the **maximum profit with the sales tax is RM4400** when 80 vases are produced.

### Summary of Results:
- **(a)** Revenue function: \( R(x) = 200x - x^2 \); revenue is maximized at \( x = 100 \).
- **(b)** Profit is maximized at \( x = 90 \) with a maximum profit of RM6100.
- **(c)** The selling price for maximum profit is RM110 per vase.
- **(d)** With a RM20 sales tax per vase, the maximum profit is RM4400, and the optimal production level is \( x = 80 \).

Would you like more details or further clarification?

Here are 5 related questions:
1. How does increasing the cost per vase affect the maximum profit?
2. What happens to the profit function if the fixed cost increases by RM500?
3. Can you determine the break-even point for the firm?
4. How would a change in the demand function impact the optimal production level?
5. What if there was a discount for bulk sales on the demand function?

**Tip**: To maximize profits, always consider both the marginal cost and marginal revenue.

A small family firm manufacturing hand-made ceramic vases knows that it has the cost function C(x) and the demand function D(x) given by C(x) = 2000 + 20x and D(x) = 200 − x where x is the number of vases manufactured and sold. Notes: Revenue, R(x) = xD(x) and Profit, P(x) = R(x) − C(x). Find (a) the revenue function and the level of production at which revenue is maximized, (b) the level of production at which profit is maximized and the value of the maximum profit, (c) the selling price in order to obtain maximum profit, (d) the maximum profit if each ceramic vase is subject to a sales tax of RM20.

Explore the step-by-step solution for maximizing revenue and profit for a small family firm producing hand-made ceramic vases. This problem involves key calculus concepts such as revenue maximization, profit maximization, and the impact of sales tax on profit. Suitable for Grades 10-12 students.

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