# Step-by-Step Solution Let's define the variables and set up the problem as a linear programming (LP) problem. ## Variables - Let \( x \) = number of **radio ads** - Let \( y \) = number of **television ads** ## Given Data - Radio ad: **$200 each**, reaches **3,000 people** - TV ad: **$500 each**, reaches **7,000 people** - Total budget: **$40,000** - At least 10 ads of each type: \( x \geq 10 \), \( y \geq 10 \) - Number of radio ads at least as great as the number of TV ads: \( x \geq y \) ## Objective **Maximize total people reached:** \[ \text{Maximize } Z = 3{,}000x + 7{,}000y \] ## Constraints 1. **Budget:** \( 200x + 500y \leq 40,000 \) 2. **Minimum ads:** \( x \geq 10 \), \( y \geq 10 \) 3. **Radio ads ≥ TV ads:** \( x \geq y \) ## Step 1: Express constraints 1. \( 200x + 500y \leq 40,000 \) 2. \( x \geq 10 \) 3. \( y \geq 10 \) 4. \( x \geq y \) ## Step 2: Find feasible region Let's use the budget constraint to see how many ads can be bought. ### Rearranging Budget Constraint Divide through by 100: \[ 2x + 5y \leq 400 \] ### Exploring possible values Since \( x \geq y \), let's try \( x = y \) (maximum possible allowed by budget): \[ 2x + 5x = 7x \leq 400 \implies x \leq \frac{400}{7} \approx 57.14 \] So, possible to have up to 57 ads of each type if \( x = y \). But is it better to use more radio or more TV? Let's check the marginal reach per dollar: - Radio: \( \frac{3000}{200} = 15 \) people per \$1 - TV: \( \frac{7000}{500} = 14 \) people per \$1 So **radio ads are slightly more cost-effective**. To maximize reach, use as many radio ads as possible. But, the constraint \( x \geq y \) means we can have more radio than TV, but not the other way around. Let's try the corner points. ## Step 3: Find Corner Points ### (A) Minimum allowed TV ads (\( y = 10 \)), maximize \( x \): \[ 2x + 5(10) \leq 400 \\ 2x + 50 \leq 400 \\ 2x \leq 350 \\ x \leq 175 \] But also, \( x \geq y = 10 \). So \( x = 175 \), \( y = 10 \) is feasible. ### (B) Minimum allowed radio ads (\( x = 10 \)), maximize \( y \): \[ 2(10) + 5y \leq 400 \\ 20 + 5y \leq 400 \\ 5y \leq 380 \\ y \leq 76 \] But also need \( x \geq y \implies 10 \geq y \implies y \leq 10 \). So only possible if \( y = 10 \). So, \( x = 10, y = 10 \) is another corner. ### (C) \( x = y \): Earlier, \( 2x + 5x = 7x \leq 400 \implies x \leq 57 \). So \( x = 57, y = 57 \). Check if all constraints satisfied: \( x \geq y \), \( x, y \geq 10 \) -- yes. ### (D) Budget line intersection with \( x = y \): Already found: \( x = y = 57 \). ## Step 4: Calculate Objective at Each Corner Point 1. **(x = 175, y = 10):** - Budget: \( 2 \times 175 + 5 \times 10 = 350 + 50 = 400 \) ✔️ - \( x \geq y \) - People reached: \( 3,000 \times 175 + 7,000 \times 10 = 525,000 + 70,000 = 595,000 \) 2. **(x = 10, y = 10):** - Budget: \( 2 \times 10 + 5 \times 10 = 20 + 50 = 70 \leq 400 \) - People reached: \( 3,000 \times 10 + 7,000 \times 10 = 30,000 + 70,000 = 100,000 \) 3. **(x = 57, y = 57):** - Budget: \( 2 \times 57 + 5 \times 57 = 114 + 285 = 399 \leq 400 \) - People reached: \( 3,000 \times 57 + 7,000 \times 57 = 171,000 + 399,000 = 570,000 \) ## Step 5: Check If Integer Solutions Are Needed Ads must be integer numbers. All above are integers. ## Step 6: Final Answer ### The best solution is: - **175 radio ads** - **10 television ads** ### Maximum people reached: \[ 3,000 \times 175 + 7,000 \times 10 = 525,000 + 70,000 = 595,000 \] --- ## **Summary Table** | Radio Ads (\(x\)) | TV Ads (\(y\)) | People Reached | Budget Used (\$) | |-------------------|----------------|----------------|------------------| | 175 | 10 | 595,000 | 40,000 | | 57 | 57 | 570,000 | 39,900 | | 10 | 10 | 100,000 | 7,000 | --- ## **Final Answer** - **Number of radio ads:** 175 - **Number of television ads:** 10 - **Maximum people reached:** 595,000 --- ## **Explanation** You should use as many radio ads as possible, since they reach more people per dollar, but you must have at least 10 TV ads and at least as many radio ads as TV ads. This strategy maximizes the number of people reached under the budget and campaign constraints.