/*! 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 21 A theater is presenting a progra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 21

A theater is presenting a program for students and their parents on drinking and driving. The proceeds will be donated to a local alcohol information center. Admission is \(\$ 2.00\) for parents and \(\$ 1.00\) for students. However, the situation has two constraints: The theater can hold no more than 150 people and every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?

Short Answer

Expert verified
To maximize the proceeds, 100 parents and 50 students should attend, raising a total of $250.

Step by step solution

01

Define the Variables

Let's define P as the number of parents and S as the number of students.
02

Set up the Objective Function

The objective is to maximize the total revenue (R). The revenue collected from the parents would be 2*P dollars and from the students it would be S dollars. So, the objective function becomes \(R = 2P + S \).
03

Set up the Constraints

The constraints given are: First, the theater can hold a maximum 150 people, i.e., \( P + S <= 150 \). Second, for every two parents, there should be at least one student. Hence, the number of students should be at least half of the number of parents, i.e., \( S >= 0.5P \). Let's convert it to \( S - 0.5P >= 0 \) which is more suitable for our solution process.
04

Apply Linear Programming Techniques

By applying linear programming methods to the objective function and constraints, we find two points of interest where constraints intersect: (0,150) and (100,50). These represent the possible maximums for the objective function.
05

Substitution and Comparison for Maximization

Substituting these points into the objective function: For (0,150), the revenue would be \( 2*0 + 150 = 150) dollars. For (100,50), the revenue would be \( 2*100 + 50 = 250 \) dollars. By comparison, we see that (100,50) gives us the maximum revenue.

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!

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

On June \(24,1948,\) the former Soviet Union blocked all land and water routes through East Germany to Berlin. A gigantic airlift was organized using American and British planes to bring food, clothing, and other supplies to the more than 2 million people in West Berlin. The cargo capacity was \(30,000\) cubic feet for an American plane and \(20,000\) cubic feet for a British plane. To break the Soviet blockade, the Western Allies had to maximize cargo capacity but were subject to the following restrictions: \(\cdot\) No more than 44 planes could be used. "The larger American planes required 16 personnel per flight, double that of the requirement for the British planes. The total number of personnel available could not exceed 512 \(\cdot\) The cost of an American flight was \(\$ 9000\) and the cost of a British flight was \(\$ 5000 .\) Total weekly costs could not exceed \(\$ 300,000\) Find the number of American and British planes that were used to maximize cargo capacity.

Determine whether cach statement makes sense or does not make sense, and explain your reasoning. Even if a linear system has a solution set involving fractions, such as \(\left\\{\left(\frac{8}{11}, \frac{43}{11}\right)\right\\}, 1\) can use graphs to determine if the solution set is reasonable.

On your next vacation, you will divide lodging between large resorts and small inns. Let \(x\) represent the number of nights spent in large resorts. Let \(y\) represent the number of nights spent in small inns. a. Write a system of inequalities that models the following conditions: You want to stay at least 5 nights. At least one night should be spent at a large resort. Large resorts average \(\$ 200\) per night and small inns average \(\$ 100\) per night. Your budget permits no more than \(\$ 700\) for lodging. b. Graph the solution set of the system of inequalities in part (a). c. Based on your graph in part (b), what is the greatest number of nights you could spend at a large resort and still stay within your budget?

The graphs of solution sets of systems of inequalities involve finding the intersection of the solution sets of two or more inequalities. By contrast, in Exercises \(71-72,\) you will be graphing the union of the solution sets of two inequalities. Graph the union of \(x-y \geq-1\) and \(5 x-2 y \leq 10\).

In Exercises 69鈥�70, rewrite each inequality in the system without absolute value bars. Then graph the rewritten system in rectangular coordinates. $$\left\\{\begin{array}{l} |x| \leq 1 \\ |y| \leq 2 \end{array}\right.$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

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.