/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Until recently hamburgers at the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 11

Until recently hamburgers at the city sports arena cost \(\$ 4\) each. The food concessionaire sold an average of 10,000 hamburgers on a game night. When the price was raised to \(\$ 4.40,\) hamburger sales dropped oft to an average of 8000 per night. (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. (b) If the concessionaire has fixed costs of \(\$ 1000\) per night and the variable cost is \(\$ .60\) per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.

Short Answer

Expert verified
Revenue-maximizing price: \(\$3\). Profit-maximizing price: \(\$3.30\).

Step by step solution

01

- Define Variables and Linear Demand Equation

Let the price of a hamburger be denoted by p and the quantity sold by q. Given two points on the demand curve: (4, 10000) and (4.40, 8000). Use the slope formula to find the slope (m) of the demand curve: \[ m = \frac{8000 - 10000}{4.40 - 4} = -5000 \] The demand equation is linear: \[ q = mp + b \] Substitute one of the points to find b: \[ 10000 = -5000(4) + b \] solving this gives \[ b = 30000 \] Therefore, the linear demand function is: \[ q = -5000p + 30000 \]
02

- Revenue Function

Revenue is given by R(p) = p * q. Substituting the demand function: \[ R(p) = p(-5000p + 30000) = -5000p^2 + 30000p \]
03

- Maximize Revenue

To find the price that maximizes revenue, take the derivative of R(p) with respect to p and set it to zero: \[ R'(p) = -10000p + 30000 \] Setting R'(p) = 0: \[ -10000p + 30000 = 0 \] Solving for p gives: \[ p = 3 \]
04

- Profit Function

Profit is given by Profit = Revenue - Costs. Where \( \text{Total Cost} = \text{Fixed Cost} + \text{Variable Cost} \). Let鈥檚 define variable (VC) and fixed costs (FC): VC = 0.60q, FC = 1000. Therefore, Cost(p) = 1000 + 0.60q. Using the demand function to substitute q in Cost(p): \[ \text{Cost}(p) = 1000 + 0.60(-5000p + 30000) = 1000 - 3000p + 18000 = -3000p + 19000 \]
05

- Maximize Profit

Profit(p) = Revenue(p) - Cost(p), substituting in values from earlier: \[ \text{Profit}(p) = (-5000p^2 + 30000p) - (-3000p + 19000) \] Simplifying gives: \[ \text{Profit}(p) = -5000p^2 + 33000p - 19000 \] Take the derivative of Profit(p) and set it to zero to find p that maximizes profit. \[ \text{Profit}'(p) = -10000p + 33000 \] Setting \(\text{Profit}'(p) = 0 \): \[ -10000p + 33000 = 0 \] Solving for p gives: \[ p = 3.30 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

linear demand curve
A linear demand curve shows the relationship between price and quantity in a straight line. Imagine two points: (4, 10000) and (4.40, 8000). The first point means that at \(4 per hamburger, 10,000 hamburgers were sold. The second indicates a reduced sale of 8,000 hamburgers when the price increased to \)4.40. We use the slope formula to find the rate of decrease:
  • Slope = \[\frac{8000 - 10000}{4.40-4} = -5000\]
This gives us a negative slope, meaning an inverse relationship between price and quantity. The linear demand equation can be specified as: \[q = -5000p + b\]. Using either point, we substitute p and q to find b: \[10000 = -5000(4) + b\]. Solving, we get b = 30000. Therefore, our demand curve is: \[q = -5000p + 30000\].
revenue function
Revenue function highlights the total income received from selling goods. Calculate it by multiplying the price per unit (p) by the quantity sold (q). With our demand curve \[q = -5000p + 30000\], the revenue formula becomes: \[R(p) = p \times q = p(-5000p + 30000)\]. Further simplifying: \[R(p) = -5000p^2 + 30000p\].
profit function
The profit function is essential for understanding earnings after costs. Profit = Revenue - Costs. Costs have two components: Fixed Costs (FC) and Variable Costs (VC). Given fixed costs of $1000 and variable costs of 0.60 per hamburger, total cost = \[1000 + 0.60q\]. Using \[q = -5000p + 30000\] to find q: \[Cost(p) = 1000 + 0.60(-5000p + 30000) = 1000 - 3000p + 18000\]. Simplifying gives: \[-3000p + 19000\]. Profit is thus: \[Profit = Revenue - Cost = (-5000p^2 + 30000p) - (-3000p + 19000) = -5000p^2 + 33000p - 19000\].
slope formula
The slope formula \[m = \frac{y2 - y1}{x2 - x1}\] calculates the rate of change between two points. Given points (4, 10000) and (4.40, 8000), the slope is: \[-5000 = \frac{8000 - 10000}{4.40 - 4}\]. This slope is crucial for forming the linear demand curve to understand how quantity changes with price.
derivative in calculus
A derivative helps find the rate of change. To maximize revenue or profit, we take the derivative of the respective function and set it to zero:
  • For Revenue: \[R(p) = -5000p^2 + 30000p\], derivative: \[R'(p) = -10000p + 30000\]. Solving \[R' = 0\]: \[p = 3\].
  • For Profit: \[Profit = -5000p^2 + 33000p - 19000\], derivative: \[Profit'(p) = -10000p + 33000\]. Solving \[Profit' = 0\]: \[p = 3.30\].
These steps tell us the optimal price points to maximize revenue and profit respectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If a total cost function is \(Q(x)=.0001 x^{3}-.06 x^{2}+12 x+100,\) is the marginal cost increasing, decreasing, or not changing at \(x=100 ?\) Find the minimum marginal cost.

A store manager wants to establish an optimal inventory policy for an item. Sales are expected to be at a steady rate and should total \(Q\) items sold during the year. Each time an order is placed a cost of \(h\) dollars is incurred. Carrying costs for the year will be \(s\) dollars per item, to be figured on the average number of items in storage during the year. Show that the total inventory cost is minimized when each order calls for \(\sqrt{2 h Q / s}\) items.

Each of the graphs of the functions has one relative maximum and one relative minimum point. Find these points using the first-derivative test. Use a variation chart as in Example 1. $$f(x)=2 x^{3}+3 x^{2}-3$$

The monthly demand equation for an electric utility company is estimated to be $$p=60-\left(10^{-5}\right) x,$$ where \(p\) is measured in dollars and \(x\) is measured in thousands of kilowatt- hours. The utility has fixed costs of 7 million dollars per month and variable costs of \(\$ 30\) per 1000 kilowatt-hours of electricity generated, so the cost function is $$a(x)=7 \cdot 10^{6}+30 x,$$ (a) Find the value of \(x\) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. (b) Suppose that rising fuel costs increase the utility's variable costs from \(\$ 30\) to \(\$ 40\), so its new cost function is $$C_{1}(x)=7 \cdot 10^{6}+40 x.$$ Should the utility pass all this increase of \(\$ 10\) per thousand kilowatt- hours on to consumers? Explain your answer.

A closed rectangular box is to be constructed with a base that is twice as long as it is wide. If the total surface area must be 27 square feet, find the dimensions of the box that will maximize the volume.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.