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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.
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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.


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 $ x_1 $ represent the number of newspaper ads.
- Let $ x_2 $ represent the number of radio ads.
- Let $ x_3 $ represent the number of TV ads.

### Objective Function:
The goal is to maximize the total number of people reached by these ads:
\[
\text{Maximize } 4000x_1 + 3000x_2 + 12000x_3
\]
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:
\[
600x_1 + 30x_2 + 3000x_3 \leq 9000
\]
2. **Newspaper ad limit**: The store can run a maximum of 30 newspaper ads:
\[
x_1 \leq 30
\]
3. **Radio ad limit**: The store can run a maximum of 60 radio ads:
\[
x_2 \leq 60
\]
4. **TV ad limit**: The store can run a maximum of 20 TV ads:
\[
x_3 \leq 20
\]
5. **Non-negativity**: The number of ads must be non-negative:
\[
x_1 \geq 0, \quad x_2 \geq 0, \quad x_3 \geq 0
\]

### 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:
\[
x_3 = \frac{9000}{3000} = 3
\]
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:
\[
\text{Exposure from TV ads} = 12000 \times 3 = 36000
\]
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.

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.

Explore how a store can maximize exposure with a $9000 advertising budget by optimizing the number of newspaper, radio, and TV ads. This linear programming problem involves maximizing exposure while staying within budget constraints. Includes detailed analysis with mathematical formulas. Suitable for Grades 11-12.

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