---
# System prepended metadata

title: '2.7 Notes on price, demand, revenue, cost, profit'
tags: [lecture, '2.7']

---

# Selling Cast Iron Pans?

In the German-style approach of breaking down concepts into their fundamental components, let's delve into the meaning and significance of each variable and function involved in our analysis:

1. **$p$:** This variable represents the price of the product, in this case, the price of a cast iron pan. It is a crucial determinant in the price-demand relationship and directly influences revenue and profit.

2. **$x$:** $x$ denotes the quantity of the product sold or produced. In our context, $x$ signifies the number of cast iron pans manufactured and sold. Understanding the relationship between quantity and price ($p$) is pivotal in determining optimal production levels for profit maximization.

3. **$R(x)$:** $R(x)$ represents the revenue function, which computes the total revenue generated from selling $x$ units of the product. It is obtained by multiplying the price per unit ($p$) by the quantity sold ($x$). Revenue is a crucial metric in assessing the financial viability and performance of a business.

4. **$C(x)$:** $C(x)$ denotes the cost function, which calculates the total cost incurred in producing $x$ units of the product. It encompasses both fixed costs (e.g., equipment, rent) and variable costs (e.g., raw materials, labor). Understanding $C(x)$ is essential for determining the profitability and cost-efficiency of production processes.

    - **Fixed Costs ($FC$):** Fixed costs are expenses that do not vary with the quantity of goods produced or sold. Examples include rent, insurance, and salaries of permanent staff. These costs remain constant regardless of the level of production or sales volume.

    - **Variable Costs per unit ($c$):** Variable costs are expenses that change in proportion to the quantity of goods produced or sold. Examples include raw materials, direct labor costs, and packaging expenses. The variable cost per unit ($c$) represents the cost incurred to produce one additional unit of the product.

5. **$P(x)$:** $P(x)$ refers to the profit function, which calculates the total profit generated from selling $x$ units of the product. It is derived by subtracting the total cost ($C(x)$) from the total revenue ($R(x)$). In essence, $P(x)$ provides insights into the financial performance of the business relative to the quantity of products sold.


## Understanding Total Cost, Total Revenue, and Total Profit Functions

In economics, total cost (TC), total revenue (TR), and total profit (TP) functions provide valuable insights into the financial performance of a business. Let's explore these functions in detail:

1. **Total Cost Function $C(x)$:**
   - **Explanation:** The total cost function represents the cumulative cost incurred in producing a certain quantity of goods or services.
   - **Example:** If $C(5) = 100$, it means that the total cost of producing 5 units of a product is $100.

2. **Total Revenue Function $R(x)$:**
   - **Explanation:** The total revenue function indicates the cumulative revenue generated from selling a specific quantity of goods or services.
   - **Example:** If $R(5) = 200$, it indicates that the total revenue from selling 5 units of a product is $200.

3. **Total Profit Function $P(x)$:**
   - **Explanation:** The total profit function represents the cumulative profit earned by a business at different production or sales levels.
   - **Example:** If $P(5) = 50$, it means that the total profit generated from selling 5 units of a product is $50.





By meticulously analyzing these fundamental components—price ($p$), quantity ($x$), revenue ($R(x)$), cost ($C(x)$), and profit ($P(x)$)—we gain comprehensive insights into the economic dynamics governing the sale and production of cast iron pans. This granular understanding enables businesses to formulate strategic decisions aimed at optimizing profitability and operational efficiency.


Here is an example where 100 pans are sold at 50 Euros each, while 80 pans are sold at 60 Euros each.


### Given:
- Two points that define the price-demand relationship: $(x_1, p_1) = (100, 50)$ and $(x_2, p_2) = (80, 60)$.
- Fixed Costs $FC = 2000$ Euros
- Variable Cost per unit $c = 30$ Euros

### Step 1: Price-Demand Equation

To determine the price-demand function $p(x)$, we use the two given points to establish the equation of the line in the form $p = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

1. Calculation of the Slope $m$:
   $$m = \frac{p_2 - p_1}{x_2 - x_1} = \frac{60 - 50}{80 - 100} = \frac{10}{-20} = -\frac{1}{2}$$

2. Determination of the y-Intercept $b$:
   Using $p = mx + b$ and one of the points:
   $$50 = -\frac{1}{2} \cdot 100 + b$$
   Hence: $$b = 50 + 50 = 100$$

Therefore, the price-demand function is:
   $$p(x) = -\frac{1}{2}x + 100$$

### Step 2: Cost Function $C(x)$

The total cost $C(x)$ consists of variable costs, which depend on the produced quantity, and fixed costs, which are independent of the production quantity.

   $$C(x) = FC + cx = 2000 + 30x$$

### Step 3: Revenue Function $R(x)$

Revenue $R(x)$ results from the product of the selling price and the sold quantity. Since the price $p(x)$ is a function of the quantity $x$, the revenue is calculated as follows:

   $$R(x) = x \cdot p(x) = x \cdot (-\frac{1}{2}x + 100)$$

### Step 4: Profit Function $P(x)$

The profit $P(x)$ is the difference between revenue and costs:

   $$P(x) = R(x) - C(x)$$
   $$P(x) = (x \cdot (-\frac{1}{2}x + 100)) - (2000 + 30x)$$

### Simplification of $R(x)$ and Calculation of $P(x)$:

   $$R(x) = -\frac{1}{2}x^2 + 100x$$

   $$P(x) = -\frac{1}{2}x^2 + 70x - 2000$$

These equations provide a comprehensive mathematical model for understanding the dynamics of price, demand, cost, revenue, and profit in the sale of cast iron pans, with the price-demand relationship explicitly defined as a function of quantity.

Given the profit function:

$$P(x) = -\frac{1}{2}x^2 + 70x - 2000$$

Let's calculate the profit for various quantities, specifically for $x = 50, 80, 100, 120,$ and $150$. This range allows us to see how profit increases or decreases with changes in production volume.

### Step 5: Profit Calculation for Various Quantities

1. **For $x = 50$ Units:**
   $$P(50) = -\frac{1}{2}(50)^2 + 70(50) - 2000$$
   Profit = 250 Euros

2. **For $x = 80$ Units:**
   $$P(80) = -\frac{1}{2}(80)^2 + 70(80) - 2000$$
   Profit = 400 Euros

3. **For $x = 100$ Units:**
   $$P(100) = -\frac{1}{2}(100)^2 + 70(100) - 2000$$
   Profit = 0 Euros

4. **For $x = 120$ Units:**
   $$P(120) = -\frac{1}{2}(120)^2 + 70(120) - 2000$$
   Profit = -800 Euros

5. **For $x = 150$ Units:**
   $$P(150) = -\frac{1}{2}(150)^2 + 70(150) - 2000$$
   Profit = -2750 Euros

Based on these values, one would expect to sell somewhere between 50-100 units to maximize profit.

Plotting $P(x)$ on a grid:




![image](https://hackmd.io/_uploads/HykttVKja.png)

The maximum profit is when 70 units are sold, so $x=70$:

\begin{align*}
P(70) &= -\frac{1}{2}(70)^2 + 70(70) - 2000 \\
&= -\frac{1}{2}(4900) + 4900 - 2000 \\
&= -\frac{2450}{2} + 4900 - 2000 \\
&= -1225 + 4900 - 2000 \\
&= 2450 - 2000 \\
&= 450 \text{ Euros}
\end{align*}


Thus the profit of selling 70 units is 450 Euros.


\begin{align*}
p(70) &= -\frac{1}{2}(70) + 100 \\
&= -35 + 100 \\
&= 65
\end{align*}

Thus the price at which to sell the 70 units is \$65 per unit.


| Quantity Sold ($x$) | Profit ($P(x)$ (Euros) | Price per Unit ($p(x)$) (Euros) | Revenue (Euros) | Cost (Euros) |
|----------------------|------------------------|----------------------------------|------------------|--------------|
| 70 (Optimal)         | 450                    | 65                               | 4550             | 4100         |


## The Power of Calculus: Using the Derivative to find the Maximum Profit
To find the maximum profit, we can take the derivative of the profit function $P(x)$ with respect to $x$, set it equal to zero, and solve for $x$. Let's proceed with this calculation:

Given the profit function:
$$P(x) = -\frac{1}{2}x^2 + 70x - 2000$$

Taking the derivative of $P(x)$ with respect to $x$:
$$P'(x) = -x + 70$$

Setting $P'(x)$ equal to zero to find the critical points:
$$-x + 70 = 0$$
$$x = 70$$

So, the critical point occurs at $x = 70$. To confirm whether this point corresponds to a maximum profit, we can perform the second derivative test.

The second derivative of $P(x)$ with respect to $x$:
$$P''(x) = -1$$

Since $P''(70) = -1 < 0$, the second derivative is negative, indicating a concave-downward shape. Therefore, at $x = 70$, we have a maximum profit.

Let's calculate the profit at $x = 70$ to determine its value.

To calculate the profit at $x = 70$, we substitute $x = 70$ into the profit function $P(x)$:

$$P(70) = -\frac{1}{2}(70)^2 + 70(70) - 2000$$

$$P(70) = -\frac{1}{2}(4900) + 4900 - 2000$$

$$P(70) = -2450 + 4900 - 2000$$

$$P(70) = 2450 - 2000$$

$$P(70) = 450 \text{ Euros}$$

Therefore, the maximum profit occurs when $x = 70$, and the profit at this point is 450 Euros.


### Maximize the following profit functions using the first and second derivative.

1. $P(x)=-x^2+53x-230$

:::spoiler

<summary> Solution: </summary>

1. To maximize the profit function $P(x) = -x^2 + 53x - 230$:

   a. First derivative:
      $$P'(x) = -2x + 53$$

   b. Setting $P'(x)$ equal to zero:
      $$-2x + 53 = 0$$
      $$x = \frac{53}{2} = 26.5$$

   c. Second derivative:
      $$P''(x) = -2$$

   Since $P''(x) = -2 < 0$, it indicates a concave-downward shape. Therefore, at $x = 26.5$, we have a maximum profit.

:::
2. $P(x)=-0.3x^2+4x-40$

:::spoiler

<summary> Solution: </summary>

2. To maximize the profit function $P(x) = -0.3x^2 + 4x - 40$:

   a. First derivative:
      $$P'(x) = -0.6x + 4$$

   b. Setting $P'(x)$ equal to zero:
      $$-0.6x + 4 = 0$$
      $$x = \frac{20}{3} \approx 6.67$$

   c. Second derivative:
      $$P''(x) = -0.6$$

   Since $P''(x) = -0.6 < 0$, it indicates a concave-downward shape. Therefore, at $x \approx 6.67$, we have a maximum profit.
:::

3. $P(x)=-2x^3+73x-800$

:::spoiler

<summary> Solution: </summary>

3. To maximize the profit function $P(x) = -2x^3 + 73x - 800$:

   a. First derivative:
      $$P'(x) = -6x^2 + 73$$

   b. Setting $P'(x)$ equal to zero:
      $$-6x^2 + 73 = 0$$
      $$x^2 = \frac{73}{6}$$
      $$x = \sqrt{\frac{73}{6}}$$

   c. Second derivative:
      $$P''(x) = -12x$$

   Since $P''(x) < 0$ for all real positive $x$, it indicates a concave-downward shape. Therefore, at $x = \sqrt{\frac{73}{6}}$, we have a maximum profit.

:::

4. $P(x)=-\dfrac{1}{4}x^2+64x-750$

:::spoiler

<summary> Solution: </summary>


4. To maximize the profit function $P(x) = -\frac{1}{4}x^2 + 64x - 750$:

   a. First derivative:
      $$P'(x) = -\frac{1}{2}x + 64$$

   b. Setting $P'(x)$ equal to zero:
      $$-\frac{1}{2}x + 64 = 0$$
      $$x = 128$$

   c. Second derivative:
      $$P''(x) = -\frac{1}{2}$$

   Since $P''(x) = -\frac{1}{2} < 0$, it indicates a concave-downward shape. Therefore, at $x = 128$, we have a maximum profit.
:::







