Problem 18 - The Calculus with Analytic Geometry

Question

A monopolist determines that if \(C(x)\) cents is the total cost of producing \(x\) units of a certain commodity, then \(C(x)=\) \(25 x+20,000\). The demand equation is \(x+50 p=5000\), where \(x\) units are demanded each week when the unit price is \(p\) cents. If the weekly profit is to be maximized, find (a) the number of units that should be produced each week, (b) the price of each unit, and (c) the weekly profit.

Step-by-Step Solution

Answer

a) 1875 units, b) 62.5 cents, c) 50312.5 cents.

Step 1: Find the Revenue Function

The demand equation is given by \[ x + 50p = 5000 \]. We need to express the price \( p \) in terms of \( x \). Solving for \( p \), we get \[ p = \frac{5000 - x}{50} \].

Step 2: Express the Revenue Function

Revenue \(R(x)\) is given by the product of the number of units sold \(x\) and the price per unit \(p\). Substituting the expression for \(p\), we get \[ R(x) = x \times \frac{5000 - x}{50} \] This simplifies to \[ R(x) = \frac{5000x - x^2}{50} \].

Step 3: Find the Profit Function

The profit function \( P(x) \) is the revenue function \( R(x) \) minus the cost function \( C(x) \). We are given that \[ C(x) = 25x + 20000 \] Therefore, the profit function is \[ P(x) = R(x) - C(x) \], i.e., \[ P(x) = \frac{5000x - x^2}{50} - (25x + 20000) \]. Simplifying, we get \[ P(x) = \frac{5000x - x^2 - 1250x - 1000000}{50} \] which simplifies to \[ P(x) = \frac{-x^2 + 3750x - 1000000}{50} \].

Step 4: Maximize the Profit

To maximize the profit, we need to find the critical points of the profit function by taking its derivative and setting it to zero. The derivative of \( P(x) \) is: \[ P'(x) = \frac{-2x + 3750}{50} = -\frac{x}{25} + 75 \]. Setting \( P'(x) = 0 \), we get \[ -\frac{x}{25} + 75 = 0 \]. Solving for \( x \), we find \[ x = 1875 \].

Step 5: Determine the Price per Unit

Substitute \( x = 1875 \) back into the price equation found in Step 1 \[ p = \frac{5000 - x}{50} \]. Thus, \[ p = \frac{5000 - 1875}{50} = \frac{3125}{50} = 62.5 \] cents.

Step 6: Calculate the Weekly Profit

Substitute \( x = 1875 \) back into the profit function: \[ P(1875) = \frac{-(1875)^2 + 3750 \times 1875 - 1000000}{50} \]. First, calculate the terms: \[ (1875)^2 = 3515625 \] \[ 3750 \times 1875 = 7031250 \]. Thus, \[ P(1875) = \frac{-3515625 + 7031250 - 1000000}{50} = \frac{3515625 - 1000000}{50} = \frac{2515625}{50} = 50312.5 \] cents.

Key Concepts

Profit Maximization

Profit maximization is a crucial concept in economics. It aims to determine the level of output that brings the most profit. In simple terms, it’s about finding the sweet spot where producing more units will bring in more revenue than it costs to make them. For a monopolist like in the exercise, the goal is to find the optimal number of units to produce to achieve maximum profit. This involves taking into account both the revenue generated from selling the products and the costs incurred in producing them. We achieve this by analyzing the profit function, which is the difference between revenue and cost.

Revenue Function

The revenue function is essential for understanding how much money a company brings in from selling its products. In our given exercise, the revenue function is derived from the demand equation, which relates the price and the quantity of units demanded. The equation is given as:

Cost Function

The cost function represents the total cost of producing a certain number of units. It usually includes both fixed and variable costs. In our exercise, the cost function is given by the equation: . Here, represents the number of units produced, is the variable cost per unit, and is the fixed cost. Fixed costs do not change with the level of output, while variable costs do. Understanding the cost function helps in determining how much it costs to produce at different levels of output, which is crucial for profit maximization.

Derivative Optimization

Derivative optimization is a mathematical technique used to find the maximum or minimum values of a function. In the context of our exercise, it involves finding the number of units that should be produced to maximize profit. We do this by taking the derivative of the profit function with respect to the number of units produced and setting it to zero. Solving the resulting equation gives us the critical points, which are potential points of maximum or minimum profit. To confirm that these points are maxima, we can use the second derivative test.

Demand Equation

The demand equation shows the relationship between the price of a product and the quantity demanded by consumers. In our exercise, the demand equation is given by: . This equation helps in understanding how the price of the product affects the number of units that consumers are willing to buy. Solving for gives us the price function, which can then be used to determine the revenue function. The demand equation is crucial for setting the optimal price to maximize profit.