Math Problem Statement

Question content area top Part 1 A store has a​ $9000 monthly advertising budget. Newspaper ads cost $ 600 each with a limit of 30 per​ month, radio ads cost $ 30 each with a limit of 60 per​ month, and TV ads cost​ $3000 each with a limit of 20 per month. About 4000 people see each newspaper​ ad, 3000 hear each radio​ ad, and​ 12,000 see each TV ad. How much of each type should be used to maximize the​ store's exposure? Interpret the results. Question content area bottom Part 1 If the store wants to maximize​ exposure, it should use    enter your response here newspaper​ ads,    enter your response here radio​ ads, and    enter your response here TV ads.

Solution

This is an optimization problem that requires maximizing the total exposure given a budget constraint. Let's break down the problem step-by-step using linear programming techniques.

Variables:

  • Let x1x_1 represent the number of newspaper ads.
  • Let x2x_2 represent the number of radio ads.
  • Let x3x_3 represent the number of TV ads.

Objective Function:

The goal is to maximize the total number of people reached by these ads: Maximize 4000x1+3000x2+12000x3\text{Maximize } 4000x_1 + 3000x_2 + 12000x_3 This is the store's total exposure in terms of how many people are reached.

Constraints:

  1. Budget constraint: The total cost of the ads cannot exceed $9000: 600x1+30x2+3000x39000600x_1 + 30x_2 + 3000x_3 \leq 9000
  2. Newspaper ad limit: The store can run a maximum of 30 newspaper ads: x130x_1 \leq 30
  3. Radio ad limit: The store can run a maximum of 60 radio ads: x260x_2 \leq 60
  4. TV ad limit: The store can run a maximum of 20 TV ads: x320x_3 \leq 20
  5. Non-negativity: The number of ads must be non-negative: x10,x20,x30x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0

Step-by-Step Solution:

Now we will use these constraints to solve for the number of ads that maximizes exposure.

1. Maximize the number of TV ads:

TV ads have the largest reach (12,000 people per ad), so it makes sense to first allocate the budget to TV ads. However, each TV ad costs $3000, and we have a $9000 budget, so the maximum number of TV ads is: x3=90003000=3x_3 = \frac{9000}{3000} = 3 This uses $9000 in the budget, which leaves no money for other ads.

2. Maximize exposure:

With all the budget spent on TV ads, the exposure will be: Exposure from TV ads=12000×3=36000\text{Exposure from TV ads} = 12000 \times 3 = 36000 Thus, the store should use 3 TV ads and no newspaper or radio ads.

Conclusion:

To maximize exposure, the store should run:

  • 0 newspaper ads
  • 0 radio ads
  • 3 TV ads

Final Answer:

If the store wants to maximize exposure, it should use:

  • 0 newspaper ads
  • 0 radio ads
  • 3 TV ads

Would you like further details or clarification on any step?

Here are 5 questions to think about next:

  1. What if the budget were increased to $15,000? How would the solution change?
  2. How would the result change if each type of ad had a different reach?
  3. What would happen if the store could run more than 20 TV ads?
  4. Could this problem be solved using other optimization methods, such as graphical methods or Excel Solver?
  5. What if the store prioritized a combination of ads instead of maximum exposure?

Tip: Linear programming problems like this one can be solved using specialized software tools like Excel Solver or Python's scipy.optimize library for more complex problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities

Formulas

Maximize exposure: 4000x_1 + 3000x_2 + 12000x_3
Budget constraint: 600x_1 + 30x_2 + 3000x_3 <= 9000
Ad limits: x_1 <= 30, x_2 <= 60, x_3 <= 20

Theorems

Linear programming constraints and objective function

Suitable Grade Level

Grades 11-12