Problem 41 - College Algebra and Calculus: An Applied Approach

Question

When the admission price for a baseball game was \(\$ 6\) per ticket, 36,000 tickets were sold. When the price was raised to \(\$ 7\), only 33,000 tickets were sold. Assume that the demand function is linear and that the variable and fixed costs for the ballpark owners are \(\$ 0.20\) and \(\$ 85,000\), respectively. (a) Find the profit \(P\) as a function of \(x\), the number of tickets sold. (b) Use a graphing utility to graph \(P\), and comment about the slopes of \(P\) when \(x=18,000\) and when \(x=36,000\). (c) Find the marginal profits when 18,000 tickets are sold and when 36,000 tickets are sold.

Step-by-Step Solution

Answer

The profit function is \(P(x) = -1/3000x^2 + 17.80x - 85000.\) The slope of the graph of this function is negative at x = 18,000 and x = 36,000. The marginal profits at these points are $5.4 and $-2 respectively.

Step 1: Formulate the Price-Ticket Relation

Given the price tickets were sold we have the following points:\n(36000, 6) and (33000, 7).\nUse these points to find the slope of the line (m) and the y-intercept (b) which represents the price function \(p(x)\).\nm = \((7 - 6) / (33000 - 36000) = - 1/3000\),\nb = \(6 - m*36000 = 18\).\nThe price function \(p(x)\) is thus defined by \(p(x) = -1/3000 x + 18\).

Step 2: Formulate the Profit Function

The profit \(P\) is given by the total revenue minus the total cost, which is \[P(x) = x*p(x) - (0.20x + 85000)\]. Replacing \(p(x)\) we get: \[P(x) = x * (-1/3000x + 18) - (0.20x + 85000)\], simplifying we get \[P(x) = -1/3000x^2 + 18x - 0.20x - 85000\]. Simplify this further to \[P(x) = -1/3000x^2 + 17.80x - 85000\].

Step 3: Graph the Profit Function

Use any graphing tool to graph this function \(P(x) = -1/3000x^2 + 17.80x - 85000\). Note the slope of the function at the points where x = 18000 and x = 36000.

Step 4: Calculate the Marginal Profits

The marginal profit is given by the derivative of the profit function. So we have \(P'(x) = -2/3000x + 17.80\).\nEvaluate P'(x) at x = 18000 and x = 36000 to get the marginal profits at these points, which turn out to be \(P'(18000) = 5.4\) and \(P'(36000) = -2\).

Key Concepts

Demand Function

In economics, the demand function describes the relationship between the price of a good and the quantity demanded by consumers. It helps us understand how price changes impact buying behavior. In our baseball game example, we use ticket prices and quantities sold to establish this relationship. The price dropped from \(7 to \)6 resulted in an increase in ticket sales from 33,000 to 36,000. This kind of data aligns with the concept of a linear demand function because the demand relationship is assumed to be linear.
Steps to Determine a Linear Demand Function:

• Identify two price-quantity pairs. Here, they are (36000, 6) and (33000, 7).
• Calculate the slope using \(m = \frac{(p_2 - p_1)}{(x_2 - x_1)}\). In this example, the slope is \(-\frac{1}{3000}\).
• Find the y-intercept using one of the points: \(b = p_1 - m \cdot x_1\). Thus, b = 18.
• The demand function, or price function, is \(p(x) = -\frac{1}{3000}x + 18\).

This linear function indicates that the sale price per ticket decreases by approximately $0.00033 for each additional ticket sold, reflecting typical consumer behavior.

Marginal Profit

Marginal profit is the extra profit earned by selling one more unit of a good. It's a critical concept in decision-making because businesses aim to maximize profit, not just total revenue. Knowing the marginal profit helps them assess the impact of adjusting sales volumes.
Calculating Marginal Profit:

• Differentiate the profit function \(P(x)\) with respect to \(x\).
• In our exercise, the profit function is \(P(x) = -\frac{1}{3000}x^2 + 17.80x - 85000\).
• Its derivative, the marginal profit function, is \(P'(x) = -\frac{2}{3000}x + 17.80\).
• Calculate the marginal profit at specific points by substituting these x-values. At x = 18000, \(P'(18000) = 5.4\), and at x = 36000, \(P'(36000) = -2\).

The positive marginal profit of $5.4 at 18,000 tickets indicates that more sales are initially beneficial, whereas a negative marginal profit at 36,000 tickets means further sales become disadvantageous, likely due to increased costs or reduced price effectiveness.

Profit Function

A profit function shows the total profit a firm earns based on various levels of output or sales. It takes into account both revenue and costs. For the baseball game, profit is defined as the difference between total revenue and total costs.
Defining the Profit Function:

• Determine total revenue using \(x \cdot p(x)\). Here, \(Revenue = x \cdot (-\frac{1}{3000}x + 18)\).
• Subtract total costs, which include variable and fixed costs. Variable cost per ticket is \(0.20, and fixed costs are \)85,000.
• Formulate the profit function: \[P(x) = x \cdot (-\frac{1}{3000}x + 18) - (0.20x + 85000)\]
• Simplify to obtain \[P(x) = -\frac{1}{3000}x^2 + 17.80x - 85000\]

Understanding this function helps the ballpark owners predict profitability at different sales volumes, making strategic decisions based on how profitability shifts with ticket sales.

Linear Function

Linear functions are foundational in economics and mathematics, representing relationships in a straight line on a graph. They are used to model situations where a change in one variable results in a consistent change in another.
Characteristics of a Linear Function:

• The equation has the form \(y = mx + b\), where \(m\) is the slope and \(b\) the y-intercept.
• The slope indicates the rate of change; in our demand function, \(-\frac{1}{3000}\) shows how the ticket price decreases as more tickets are sold.
• The y-intercept represents the starting value when the independent variable is zero.
• In our context, the linearity assumption simplifies calculating anticipated revenues or adjusting sales strategies, as changes are predictable.

Linear functions are vital to economic modeling for their simplicity and ease of visual representation, assisting decision-makers in forecasting scenarios over a feasible range of values.