Marginal Revenue, Average Cost, Profit, Price & Demand Function - Calculus

The video begins by focusing on economic problems such as marginal cost, revenue, and profit. It introduces the cost function (c(x)) as the total cost to produce 'x' items. The average cost is then defined as the cost function divided by 'x'. The concept of marginal cost as the first derivative of the cost function is also explained. The video aims to help viewers understand how to find the optimal number of units to produce to minimize the average cost per unit, which is crucial for business efficiency and profitability.

The script provides a detailed example to illustrate the calculation of average and marginal costs. Given a cost function c(x) = 5000 + 15x + 0.01x^2, the total cost to produce 100 units is calculated, resulting in $6600. The average cost per unit is then determined by dividing the total cost by the number of units, yielding an average cost of $66 per unit. The marginal cost function is derived and used to calculate the additional cost of producing one more unit beyond 100, which is found to be approximately $17 per unit.

The video explains how to find the production level that minimizes the average cost. It is shown that the average cost is minimized when the average cost function is equal to the marginal cost. The script guides viewers through setting these two functions equal to each other and solving for 'x'. Using the cost function from the previous example, the minimum average cost is found to occur at a production level of 707 units, with a minimum average cost of $29.14 per unit.

To validate the minimum average cost, the video compares the average cost per unit at different production levels. It shows that the average cost at 100 units is $66, and it is higher at 600 units ($29.33) and 800 units ($29.25) compared to the minimum average cost at 707 units ($29.14). This comparison confirms that producing 707 units results in the lowest average cost, which is essential for maximizing profit.

The video presents a new cost function, c(x) = 30,000 + 200x + 0.1x^2, and repeats the process of finding the cost, average cost, and marginal cost for producing 100 units. The total cost for 100 units is calculated to be $51,000, resulting in an average cost of $510 per unit. The marginal cost at 100 units is determined to be $220 per unit. The video encourages viewers to practice by finding the production level that minimizes the average cost and calculating the minimum average cost for this new function.

The script explains the process of setting the marginal cost equal to the average cost to find the production level that minimizes the average cost. By solving the equation 0.1x = 30,000/x, the optimal production level is found to be 548 units. The minimum average cost at this production level is calculated to be $309.54 per unit. The video also demonstrates an alternative method of finding the minimum average cost by taking the derivative of the average cost function and setting it to zero.

The video analyzes the relationship between average cost, marginal cost, and production levels. It shows that at a production level of 100 units, the average cost is $510, and at 548 units, it is minimized at $309.54. At 1000 units, the average cost increases slightly to $330. The marginal cost is calculated at different production levels, and it is observed that at the minimum average cost, the average cost function equals the marginal cost. The video explains that comparing the average cost to the marginal cost can indicate whether to increase or decrease production to minimize the average cost.

The video introduces the price function (p(x)) or demand function, which represents the price per unit that a company can charge for selling 'x' units. It explains the inverse relationship between supply (x) and demand, where an increase in supply leads to a decrease in demand and price, and vice versa. Examples of calculators and diamonds are used to illustrate this concept. The video emphasizes the importance of understanding the relationship between supply, demand, and price.

The script defines the revenue function (r(x)) as the product of the number of units sold (x) and the price function (p(x)). It provides examples to illustrate how revenue is calculated and relates it to the concept of profit, which is the difference between revenue and cost. The video explains that to maximize profit, one must consider both revenue and cost, and it introduces the profit function (P(x)) as revenue minus cost. It also touches on the concept of marginal profit as the first derivative of the profit function.

The video concludes with examples of how to find the production level that maximizes profit. Given cost and demand functions, it shows how to calculate the revenue function, find the profit function, and then determine the production level at which profit is maximized. The process involves setting the first derivative of the profit function to zero and solving for 'x', or alternatively, setting the marginal revenue equal to the marginal cost and solving for 'x'. The video provides calculations for a scenario with a constant price function and another with a variable price function, demonstrating how to find the production level and maximum profit in each case.

The video wraps up by summarizing the process for finding the production level that maximizes profit and calculating the maximum profit. It emphasizes the importance of understanding cost functions, revenue functions, and the relationship between marginal revenue and marginal cost. The video thanks viewers for watching and encourages them to apply these concepts to real-world business scenarios.