Problem 42 - College Algebra and Calculus: An Applied Approach

Question

The profit \(P\) made by a cinema from selling \(x\) bags of popcorn can be modeled by \(P=2.36 x-\frac{x^{2}}{25,000}-3500, \quad 0 \leq x \leq 50,000\) (a) Find the intervals on which \(P\) is increasing and decreasing. (b) If you owned the cinema, what price would you charge to obtain a maximum profit for popcorn? Explain your reasoning.

Step-by-Step Solution

Answer

The profit function is increasing when \(0<=x<29500\) and decreasing when \(29500

Step 1: Find the First Derivative

The first derivative of a function shows where the function increases or decreases. Differentiating \(P=2.36 x-\frac{x^{2}}{25,000}-3500\) gives \(P'=2.36-\frac{x}{12500}\)

Step 2: Find Critical points

Critical points occur where the first derivative is zero or undefined. Equate \(P'\) to zero and solve for \(x\): \(2.36-\frac{x}{12500}=0=>x=29500\) . \(P'\) is decreasing, \(P\) has a maximum at \(x=29500\), because the maximum value of \(x\) is 50000.

Step 3: Analyze intervals

With the critical point \(x=29500\), we have two intervals to consider: \(0 \leq x < 29500\) and \(29500 < x \leq 50000\). For the first interval, choose a test point in \(0 \leq x < 29500\), say \(x=10000\). Substituting into \(P'\) yields a positive result, which means \(P\) is increasing in this interval. For the second interval, choose a test point in \(29500 < x \leq 50000\), say \(x=40000\). Substituting into \(P'\) yields a negative result, indicating that \(P\) is decreasing in this interval.

Step 4: Find the price for maximum profit

As the price for the popcorn bag is not given, we cannot determine an exact price for maximum profit. However, we know that at \(x=29500\) bags sold, the profit is maximized. If necessary information was given, we could divide the total revenue (profit + costs) by the number of bags to find the optimal price.

Key Concepts

First Derivative Analysis

When we speak about first derivative analysis in calculus, we're looking at a tool that provides important insights into the behavior of functions. Specifically, in the realm of economics, understanding how profit changes with respect to the number of goods sold is crucial.

To perform a first derivative analysis, we differentiate the function that represents profit with respect to the number of items, which in this case, is bags of popcorn. Mathematically, we obtain the derivative by applying differentiation rules. The first derivative of the profit function, denoted as \( P' \), indicates the rate of change of profit with respect to the number of bags sold. In simple terms, it tells us if the profit is increasing or decreasing as more popcorn bags are sold.

Positive values of \( P' \) indicate an increasing profit, while negative values suggest it’s decreasing. By finding where \( P' \) is zero, we can locate critical points that potentially correspond to maximum or minimum profit amounts. As we can observe in the solution, when \( P' = 2.36 - \frac{x}{12500} \), it shows the rate at which profit increases or decreases for different values of \( x \).

Critical Points in Functions

The concept of critical points is pivotal when analyzing functions, especially in the context of economic decisions like setting prices to optimize profit. Critical points are found where the first derivative equals zero or does not exist.

In our popcorn sales example, setting \( P' = 0 \) and solving for \( x \) gives us the number of bags sold that could potentially maximize profit. In this case, selling \( x = 29500 \) bags of popcorn is a critical point because the first derivative changes sign around this point.

Why Critical Points Matter

For business applications, a critical point can indicate a peak profit (a maximum) or the lowest profit (a minimum). By testing the intervals around \( x = 29500 \), we identified that the profit function increases before reaching this critical point and decreases thereafter. This kind of analysis adds depth to decision-making processes because it tells the business owner, quite precisely, the optimal sales volume to maximize profit.

Optimization of Profit

The ultimate goal for many businesses is to optimize profit, which means finding a balance between costs and revenue to achieve the highest possible profit.

In calculus, optimization involves using derivative tests to locate maximum and minimum values of profit functions. As per the worked exercise, after finding the critical point and analyzing intervals, we've determined that at \( x = 29500 \), we reach a maximum profit. The role of optimization extends beyond identifying this point—it requires understanding what these quantities mean in real-world units.

Applying Optimization

For instance, knowing that 29,500 bags equate to maximum profit doesn't tell us how much to charge for each bag. To fine-tune the optimization, we need to factor in the cost of making and selling each popcorn bag, market demand, and competitive pricing.

Once all relevant economic factors are considered, a business can set an optimal selling price. While the calculation in the textbook didn't provide enough data to determine the exact price, the methodology demonstrated is a powerful tool in the economics of decision-making and business strategy formulation.

Previous

Problem 42

Next

Problem 42

Other exercises in this chapter

Problem 41

In Exercises, find the maximum value of \(\left|f^{\prime \prime}(x)\right|\) on the closed interval. (You will use this skill in Section \(12.4\) to estimate t

View solution

Problem 42

In Exercises, use a graphing utility to graph the function and identify all relative extrema and points of inflection. $$ f(x)=x^{3}-\frac{3}{2} x^{2}-6 x $$

View solution

Problem 42

Demand In Exercises, find the rate of change of \(x\) with respect to \(p\). $$ p=\frac{4}{0.000001 x^{2}+0.05 x+1} \quad x \geq 0 $$

View solution

Problem 42

In Exercises, find the second derivative and solve the equation \(f^{\prime \prime}(x)=0\). $$ f(x)=\frac{x}{x-1} $$

View solution

Key Concepts

First Derivative Analysis

When we speak about first derivative analysis in calculus, we're looking at a tool that provides important insights into the behavior of functions. Specifically, in the realm of economics, understanding how profit changes with respect to the number of goods sold is crucial.

To perform a first derivative analysis, we differentiate the function that represents profit with respect to the number of items, which in this case, is bags of popcorn. Mathematically, we obtain the derivative by applying differentiation rules. The first derivative of the profit function, denoted as \( P' \), indicates the rate of change of profit with respect to the number of bags sold. In simple terms, it tells us if the profit is increasing or decreasing as more popcorn bags are sold.

Positive values of \( P' \) indicate an increasing profit, while negative values suggest it’s decreasing. By finding where \( P' \) is zero, we can locate critical points that potentially correspond to maximum or minimum profit amounts. As we can observe in the solution, when \( P' = 2.36 - \frac{x}{12500} \), it shows the rate at which profit increases or decreases for different values of \( x \).

Critical Points in Functions

The concept of critical points is pivotal when analyzing functions, especially in the context of economic decisions like setting prices to optimize profit. Critical points are found where the first derivative equals zero or does not exist.

In our popcorn sales example, setting \( P' = 0 \) and solving for \( x \) gives us the number of bags sold that could potentially maximize profit. In this case, selling \( x = 29500 \) bags of popcorn is a critical point because the first derivative changes sign around this point.

Why Critical Points Matter

For business applications, a critical point can indicate a peak profit (a maximum) or the lowest profit (a minimum). By testing the intervals around \( x = 29500 \), we identified that the profit function increases before reaching this critical point and decreases thereafter. This kind of analysis adds depth to decision-making processes because it tells the business owner, quite precisely, the optimal sales volume to maximize profit.

Optimization of Profit

The ultimate goal for many businesses is to optimize profit, which means finding a balance between costs and revenue to achieve the highest possible profit.

In calculus, optimization involves using derivative tests to locate maximum and minimum values of profit functions. As per the worked exercise, after finding the critical point and analyzing intervals, we've determined that at \( x = 29500 \), we reach a maximum profit. The role of optimization extends beyond identifying this point—it requires understanding what these quantities mean in real-world units.

Applying Optimization

For instance, knowing that 29,500 bags equate to maximum profit doesn't tell us how much to charge for each bag. To fine-tune the optimization, we need to factor in the cost of making and selling each popcorn bag, market demand, and competitive pricing.

Once all relevant economic factors are considered, a business can set an optimal selling price. While the calculation in the textbook didn't provide enough data to determine the exact price, the methodology demonstrated is a powerful tool in the economics of decision-making and business strategy formulation.