Problem 20 - Calculus and its applications

Question

Rad Designs sells two kinds of sweatshirts that compete with one another. Their demand functions are expressed by the following relationships: $$ \begin{array}{l} q_{1}=78-6 p_{1}-3 p_{2} \\ q_{2}=66-3 p_{1}-6 p_{2} \end{array} $$ where \(p_{1}\) and \(p_{2}\) are the prices of the sweatshirts, in multiples of \(\$ 10,\) and \(q_{1}\) and \(q_{2}\) are the quantities of the sweatshirts demanded, in hundreds of units. a) Find a formula for the total-revenue function, \(R,\) in terms of the variables \(p_{1}\) and \(p_{2}\). [Hint: \(R=p_{1} q_{1}+p_{2} q_{2} ;\) then substitute expressions from equations (1) and (2) to find \(R\left(p_{1}, p_{2}\right)\). ] b) What prices \(p_{1}\) and \(p_{2}\) should be charged for each product in order to maximize total revenue? c) How many units will be demanded? d) What is the maximum total revenue?

Step-by-Step Solution

Answer

Maximize revenue by setting \(p_1 = 4.5\) and \(p_2 = 3.5\), demanding 33.5 and 28.5 thousand units respectively, achieving maximum revenue of 250.5 thousand dollars.

Step 1: Define the Total Revenue Function

The total revenue function \(R\) is defined as: \[ R = p_1 q_1 + p_2 q_2 \] where \(p_1\) and \(p_2\) are the prices, and \(q_1\) and \(q_2\) are the quantities demanded. We need to express \(R\) in terms of \(p_1\) and \(p_2\).

Step 2: Substitute the Demand Functions

Substitute the given demand functions into the revenue equation: \[ q_1 = 78 - 6p_1 - 3p_2 \] \[ q_2 = 66 - 3p_1 - 6p_2 \] Thus, \[ R = p_1(78 - 6p_1 - 3p_2) + p_2(66 - 3p_1 - 6p_2) \]

Step 3: Expand the Total Revenue Equation

Expand the equation by multiplying out the terms: \[ R = 78p_1 - 6p_1^2 - 3p_1p_2 + 66p_2 - 3p_2p_1 - 6p_2^2 \] Simplify and combine like terms: \[ R = -6p_1^2 - 6p_2^2 - 6p_1p_2 + 78p_1 + 66p_2 \]

Step 4: Calculate the Partial Derivatives

Find the partial derivatives of \(R\) with respect to \(p_1\) and \(p_2\) to locate the critical points: \[ \frac{\partial R}{\partial p_1} = -12p_1 - 6p_2 + 78 = 0 \] \[ \frac{\partial R}{\partial p_2} = -6p_1 - 12p_2 + 66 = 0 \]

Step 5: Solve the System of Equations

Solve the system of equations to find \(p_1\) and \(p_2\): \[ -12p_1 - 6p_2 + 78 = 0 \] \[ -6p_1 - 12p_2 + 66 = 0 \] Simplify the system: \[ -12p_1 - 6p_2 = -78 \] \[ -6p_1 - 12p_2 = -66 \] Solve for \(p_1 = 4.5\) and \(p_2 = 3.5\).

Step 6: Calculate the Demand

With \(p_1 = 4.5\) and \(p_2 = 3.5\), substitute back into the demand functions to calculate \(q_1\) and \(q_2\): \[ q_1 = 78 - 6(4.5) - 3(3.5) = 33.5 \text{ (thousands)} \] \[ q_2 = 66 - 3(4.5) - 6(3.5) = 28.5 \text{ (thousands)} \]

Step 7: Calculate the Maximum Total Revenue

Substitute \(p_1 = 4.5\), \(p_2 = 3.5\), \(q_1 = 33.5\), and \(q_2 = 28.5\) into the total revenue equation: \[ R = 4.5 \times 33.5 + 3.5 \times 28.5 = 150.75 + 99.75 = 250.5 \text{ (thousands)} \]

Key Concepts

Demand Functions

Demand functions are essential tools in economics that describe how the quantity demanded of a product relates to various factors, primarily its price. In our exercise, the demand functions for two types of sweatshirts are given as:

• \( q_1 = 78 - 6p_1 - 3p_2 \)
• \( q_2 = 66 - 3p_1 - 6p_2 \)

Here, \( q_1 \) and \( q_2 \) represent the quantities demanded, while \( p_1 \) and \( p_2 \) are the prices set for each sweatshirt. These equations tell us that as price \( p_1 \) increases, the demand for the first sweatshirt \( q_1 \) decreases. Likewise, an increase in \( p_2 \) leads to a decrease in \( q_2 \).
Understanding these relationships is vital because they help retailers decide how to price their products effectively to meet desired demand levels.

Partial Derivatives

In calculus, a partial derivative represents the rate at which a function changes with respect to one of its variables, while keeping the other variables constant. In the context of maximizing revenue, we use the revenue function that has been formulated in terms of \( p_1 \) and \( p_2 \): \[ R = -6p_1^2 - 6p_2^2 - 6p_1p_2 + 78p_1 + 66p_2 \]
To find the critical points, we calculate the partial derivatives of \( R \) concerning each price variable:

• \( \frac{\partial R}{\partial p_1} \) tells us how the revenue changes when \( p_1 \) changes, holding \( p_2 \) constant.
• \( \frac{\partial R}{\partial p_2} \) shows the change in revenue with a change in \( p_2 \), keeping \( p_1 \) constant.

These derivatives allow us to determine the optimum price points where the revenue is maximized, helping businesses to make informed pricing decisions.

System of Equations

Solving a system of equations is a mathematical approach used to find values for variables that satisfy all equations involved. Our revenue maximization problem leads to a system of two equations:

• \( -12p_1 - 6p_2 + 78 = 0 \)
• \( -6p_1 - 12p_2 + 66 = 0 \)

These equations result from setting the partial derivatives of the revenue function equal to zero, aiming to find the critical points.
Solving this system involves techniques such as substitution or elimination to find the values of \( p_1 \) and \( p_2 \) that maximize revenue. In this case, the solution \( p_1 = 4.5 \) and \( p_2 = 3.5 \) represents the optimal pricing strategy for the products to achieve maximum revenue.

Maximizing Revenue

Maximizing revenue is a common goal in business and involves adjusting various factors to achieve the highest possible sales income. In our case, we achieve this by finding the right price points \( p_1 \) and \( p_2 \) that maximize the total revenue function \( R \).
To do this, we first express the revenue function in terms of demand functions, expand and simplify it. We then calculate the partial derivatives to find critical points and solve the system of equations to find optimal prices.

• The result, \( p_1 = 4.5 \) and \( p_2 = 3.5 \), shows the prices that create equilibrium between demand and sales potential.
• Substituting these back into the demand functions gives us the corresponding quantities demanded, ensuring those prices are supported by market demand.

Finally, substituting these values into the revenue equation, we confirm that the maximum revenue possible under these conditions is \( 250.5 \) thousand dollars. This calculated maximum provides strategic insights for business decision-making.