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 represent the number of newspaper ads.
- Let represent the number of radio ads.
- Let represent the number of TV ads.
Objective Function:
The goal is to maximize the total number of people reached by these ads: This is the store's total exposure in terms of how many people are reached.
Constraints:
- Budget constraint: The total cost of the ads cannot exceed $9000:
- Newspaper ad limit: The store can run a maximum of 30 newspaper ads:
- Radio ad limit: The store can run a maximum of 60 radio ads:
- TV ad limit: The store can run a maximum of 20 TV ads:
- Non-negativity: The number of ads must be non-negative:
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: 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: 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:
- What if the budget were increased to $15,000? How would the solution change?
- How would the result change if each type of ad had a different reach?
- What would happen if the store could run more than 20 TV ads?
- Could this problem be solved using other optimization methods, such as graphical methods or Excel Solver?
- 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.
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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
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