Let's break down and solve the problem step by step. --- ## **Given Data** | Constraint | Television | Radio | Online | |-----------------------------|--------------|-------------|-------------| | Audience per advertisement | 100,000 | 18,000 | 40,000 | | Cost per advertisement | $2,000 | $300 | $600 | | Maximum media usage | 10 | 20 | 10 | **Additional constraints:** - **Radio** ads ≤ 50% of total ads. - **Television** ads ≥ 10% of total ads. Let: - \( x_1 \) = number of TV ads - \( x_2 \) = number of Radio ads - \( x_3 \) = number of Online ads --- ## **a. Formulate the Linear Program** ### **Objective:** Maximize total audience contact: \[ \text{Maximize } Z = 100{,}000x_1 + 18{,}000x_2 + 40{,}000x_3 \] ### **Constraints:** 1. **Budget:** \[ 200x_1 + 300x_2 + 600x_3 \leq 18200 \] 2. **Maximum Usage:** \[ x_1 \leq 10,\quad x_2 \leq 20,\quad x_3 \leq 10 \] 3. **Radio ≤ 50% of total ads:** \[ x_2 \leq .5(x_1 + x_2 + x_3) \implies x_2 \leq x_1 + x_3 \] 4. **Television ≥ 10% of total ads:** \[ x_1 \geq .1(x_1 + x_2 + x_3) \implies .9x_1 \geq .1x_2 + .1x_3 \implies 9x_1 \geq x_2 + x_3 \] 5. **Non-negativity:** \[ x_1, x_2, x_3 \geq \] (and must be integers) --- ## **b. Solution Steps** Let's try to maximize \( x_1 \) because it gives the highest audience per dollar: ### **Step 1: Maximize Television Ads** - Max TV ads = 10 (from constraint). - Cost: \( 10 \times \$200 = \$20,000 \) (exceeds budget) - So, let's try \( x_1 = 9 \): \( 9 \times \$200 = \$18,000 \) - Remaining budget: \( \$18,200 - \$18,000 = \$200 \) But $200 is not enough for any radio or online ad. Let's check the TV lower for more flexibility. #### Try \( x_1 = 8 \): - TV cost: \( 8 \times \$200 = \$16,000 \) - Remaining: \$2,200 Let's maximize online ads (higher audience per dollar than radio): - Each online ad = \$600 → Max possible: \( \left\lfloor \frac{220}{600} \right\rfloor = 3 \) - Cost: \( 3 \times \$600 = \$1,800 \) - Remaining: \$400 Use the rest for radio: \( \frac{400}{300} = 1.33 \) → 1 ad possible - Cost for 1 radio ad = \$300 - Leftover: \$100 (can't use) So, possible allocation: \( x_1 = 8, x_2 = 1, x_3 = 3 \) #### Check constraints: - **Total ads:** \( 8 + 1 + 3 = 12 \) - **Radio ≤ 50%:** \( x_2 = 1 \leq .5 \times 12 = 6 \) ✔️ - **TV ≥ 10%:** \( x_1 = 8 \geq .1 \times 12 = 1.2 \) ✔️ - **Max usage:** All within limits #### Objective value: \[ Z = 100,000 \times 8 + 18,000 \times 1 + 40,000 \times 3 = 800,000 + 18,000 + 120,000 = \boxed{938,000} \] #### Check if we can do better with fewer TV ads and more online/radio for higher total audience --- ### **Step 2: Try maximizing online ads (since \$/audience is next best)** - Max online ads: 10 (from constraint) - Cost: \( 10 \times \$600 = \$6,000 \) - Remaining budget: \( \$18,200 - \$6,000 = \$12,200 \) - Max TV ads with this: \( \left\lfloor \frac{12,200}{2,000} \right\rfloor = 6 \) - Cost: \( 6 \times \$2,000 = \$12,000 \) - Remaining: \$200 (enough for radio ads) Total ads: \( 10 + 6 = 16 \) - **Radio ≤ 50%:** ≤ 8 ✔️ - **TV ≥ 10%:** \( 6 \geq .1 \times 16 = 1.6 \) ✔️ Objective: \( 40,000 \times 10 + 100,000 \times 6 = 400,000 + 600,000 = 1,000,000 \) **This is better!** --- #### Check if we can add any radio ads if we do fewer online ads. Try 9 online ads (\$5,400), TV: \( \left\lfloor \frac{18,200-5,400}{2,000} \right\rfloor = 6.4 \) → 6 TV ads (\$12,000), left: \$800, radio: \( \left\lfloor \frac{800}{300} \right\rfloor = 2 \), cost: \$600, left: \$200 Total: 9 online, 6 TV 2 radio = 17 ads - **Radio ≤ 50%:** 2 ≤ 8.5 ✔️ - **TV ≥ 10%:** 6 ≥ 1.7 ✔️ Objective: \( 100,000 \times 6 + 18,000 \times 2 + 40,000 \times 9 = 600,000 + 36,000 + 360,000 = 996,000 \) < 1,000,000 --- ### **Step 3: Try maximizing TV ads to 7** - TV: 7 (\$14,000), left: \$4,200 - Online: \( \left\lfloor \frac{4,200}{600} \right\rfloor = 7 \) (\$4,200) - Total ads: 14 Objective: \( 7 \times 100,000 + 7 \times 40,000 = 700,000 + 280,000 = 980,000 \) Lower than 1,000,000. --- ### **Step 4: Try all online ads and rest radio** - Online: 10 (\$6,000), left: \$12,200 - Radio: \( \left\lfloor \frac{12,200}{300} \right\rfloor = 40 \), but max is 20, so use 20 (\$6,000) - Left: \$6,200 - TV: \( \left\lfloor \frac{6,200}{2,000} \right\rfloor = 3 \) (\$6,000) - Left: \$200 Total: 10 online, 20 radio, 3 TV = 33 ads Check radio ≤ 50%: 20 ≤ 16.5? No! Violates constraint. --- ### **Step 5: Try balancing TV and online for more radio** But radio has lowest audience per dollar, so adding more radio doesn't help maximize audience. --- ## **Final Solution** - **TV ads: 6** - **Online ads: 10** - **Radio ads: ** - **Total cost:** \( 6 \times \$2,000 + 10 \times \$600 = \$12,000 + \$6,000 = \$18,000 \) - **Unused:** \$200 (not enough for any more ads) - **Total audience contact:** \( 100,000 \times 6 + 40,000 \times 10 = 600,000 + 400,000 = \boxed{1,000,000} \) **All constraints are satisfied.** --- ### **Final Answer (a)** > **To maximize audience contact with a budget of $18,200:** > - Number of TV ads: **6** > - Number of Radio ads: **** > - Number of Online ads: **10** > - **Total audience contact:** **1,000,000** > - **Budget allocation:** $12,000 (TV), $ (Radio), $6,000 (Online) --- ## **b. If an extra $100 were allocated** New budget: $18,300. Still not enough for another online ($600), TV ($2,000), or radio ($300) ad. So, the **audience contact would not increase** with an extra $100, because it is not enough to purchase any additional ad. --- ### **Final Answer (b)** > **Audience contact would not increase** if an extra $100 were allocated to the promotional budget, because it is not sufficient to purchase an additional advertisement in any medium. --- ### **Summary Table** | Medium | Ads Run | Cost | Audience Reached | |-------------|---------|-----------|-------------------| | Television | 6 | $12,000 | 600,000 | | Radio | | $ | | | Online | 10 | $6,000 | 400,000 | | **Total** | **16** | **$18,000**| **1,000,000** |