Question

The monthly demand function and cost function for \(x\) newspapers at a newsstand are given by \(p=5-0.001 x\) and \(C=35+1.5 x\) (a) Find the monthly revenue \(R\) as a function of \(x\). (b) Find the monthly profit \(P\) as a function of \(x\). (c) Complete the table. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 600 & 1200 & 1800 & 2400 & 3000 \\ \hline d R / d x & & & & & \\ \hline d P / d x & & & & & \\ \hline P & & & & & \\ \hline \end{array} $$

Short answer

The revenue function is \(R = 5x - 0.001x^2\), and the profit function is \(P = 3.5x - 0.001x^2 - 35\). After completing the table, the values obtained are: For \(x = 600\), \(dR/dx = 4.8\), \(dP/dx = 3.3\), \(P = 2050\), For \(x = 1200\), \(dR/dx = 4.6\), \(dP/dx = 3.1\), \(P = 3700\),For \(x = 1800\), \(dR/dx = 4.4\), \(dP/dx = 2.9\), \(P = 5050\), For \(x = 2400\), \(dR/dx = 4.2\), \(dP/dx = 2.7\), \(P = 6100\), For \(x = 3000\), \(dR/dx = 4.0\), \(dP/dx = 2.5\), \(P = 6850\)

Step-by-step solution

Determine the Revenue Function

Given that the price \(p\) per newspaper and the number of newspapers \(x\) form the revenue function, you can find the revenue \(R\) function by multiplying \(p\) and \(x\). The monthly revenue \(R\) is expressed as \(R = p \cdot x = (5 - 0.001x) \cdot x = 5x - 0.001x^2\).

Determine the Profit Function

The profit \(P\) is obtained by subtracting the cost \(C\) from the revenue \(R\). So, the monthly profit \(P\) for the newsstand is expressed as \(P = R - C = (5x - 0.001x^2) - (35 + 1.5x) = 3.5x - 0.001x^2 - 35\).

Derive the Revenue and Profit Functions

To complete the table, we must find the derivatives of the revenue and profit functions. The derivative of the revenue function \(R\) with respect to \(x\) is \(dR/dx = 5 - 0.002x\). The derivative of the profit function \(P\) with respect to \(x\) is \(dP/dx = 3.5 - 0.002x\).

Calculate Values for the Table

Substitute the given values of \(x\) into \(dR/dx\), \(dP/dx\), and \(P\) to complete the table. This results in: For \(x = 600\), \(dR/dx = 4.8\), \(dP/dx = 3.3\), \(P = 2050\), For \(x = 1200\), \(dR/dx = 4.6\), \(dP/dx = 3.1\), \(P = 3700\), For \(x = 1800\), \(dR/dx = 4.4\), \(dP/dx = 2.9\), \(P = 5050\), For \(x = 2400\), \(dR/dx = 4.2\), \(dP/dx = 2.7\), \(P = 6100\), For \(x = 3000\), \(dR/dx = 4.0\), \(dP/dx = 2.5\), \(P = 6850\)

Key concepts

Demand Function

The demand function represents the relationship between the quantity of goods demanded and their price. Typically, it's expressed as an equation showing how price ( p generally decreases as quantity ( x ) increases. In our example, the demand function for the newspapers is offered as p=5-0.001x , which suggests:

• As more newspapers are desired (x increases), the price per newspaper falls.
• This negative relationship reflects a common economic principle: the law of demand.

Understanding demand functions helps businesses set optimal prices and adjust to market changes.

Cost Function

Cost functions illustrate the total cost incurred by a business in producing a certain number of goods or services. In this scenario, the cost function is given by C=35+1.5x . Here's what it means:

• The constant term (35) represents fixed costs, which are expenses that don't change with production levels like rent or salaries.
• The term 1.5x represents variable costs, which increase with the number of newspapers produced (e.g., ink and paper).

Knowing your cost function allows you to predict how costs will change with different levels of production, crucial for financial planning.

Derivatives in Economics

Derivatives in economics help in analyzing how functions like revenue and profit respond to changes in variables. They are essential for:

• Determining the rate of change - the derivative of the revenue function, \(dR/dx = 5 - 0.002x\), indicates how revenue changes as more newspapers are sold.
• Understanding the slope - it helps in identifying increasing or decreasing tendencies in economic functions.

By calculating these derivatives, a business can make informed decisions about how best to increase revenue or profit. For instance, it can suggest where to set production levels for maximal efficiency.

Optimization in Business

Optimization plays a crucial role in business, as it involves finding the best possible solution under given constraints. In the context of this exercise:

• Optimization concerns identifying the number of newspapers to produce to maximize profit.
• Using derivatives like \(dP/dx = 3.5 - 0.002x\) for profit, businesses can discover when adding an additional newspaper starts to decrease the profit.

Employing these mathematical tools provides businesses with strategies to optimize performance across various operational facets, such as pricing or production levels, and ensures resources are used most efficiently for the highest gain.