1. **Problem Statement:** KIA Motors Limited (KML) wants to maximize total sales by sending promotions to two customer groups: current customers and new customers. 2. **Define Decision Variables:** Let $x$ = number of promotions sent to current customers. Let $y$ = number of promotions sent to new customers. 3. **Objective Function:** Maximize total sales = sales from current customers + sales from new customers. Sales from current customers = promotions sent ($x$) \times response rate (25%) \times sales rate (12%) = $x \times 0.25 \times 0.12 = 0.03x$. Sales from new customers = $y \times 0.20 \times 0.20 = 0.04y$. So, maximize $$Z = 0.03x + 0.04y$$ 4. **Constraints:** - Minimum test drives for current customers: $0.25x \geq 30000$ or $x \geq 120000$. - Minimum test drives for new customers: $0.20y \geq 10000$ or $y \geq 50000$. - Current customers test drives at least twice new customers: $0.25x \geq 2 \times 0.20y$ or $0.25x \geq 0.40y$ or $x \geq 1.6y$. - Budget constraint: $4x + 6y \leq 1200000$. - Non-negativity: $x \geq 0$, $y \geq 0$. 5. **Standard Form:** Maximize $$Z = 0.03x + 0.04y$$ Subject to: $$x - 120000 \geq 0$$ $$y - 50000 \geq 0$$ $$x - 1.6y \geq 0$$ $$4x + 6y \leq 1200000$$ $$x, y \geq 0$$ 6. **Graphical Solution:** Plot constraints and find feasible region. - From $x \geq 120000$ and $y \geq 50000$, feasible region is right and above these lines. - From $x \geq 1.6y$, region is above the line $x=1.6y$. - Budget line: $4x + 6y = 1200000$ or $y = \frac{1200000 - 4x}{6}$. Check corner points: - Point A: $(120000, 50000)$ - Point B: Intersection of $x=1.6y$ and $4x + 6y = 1200000$. Solve for B: $$x=1.6y$$ $$4(1.6y) + 6y = 1200000$$ $$6.4y + 6y = 1200000$$ $$12.4y = 1200000$$ $$y = 96774.19$$ $$x = 1.6 \times 96774.19 = 154838.71$$ Calculate objective at A: $$Z_A = 0.03(120000) + 0.04(50000) = 3600 + 2000 = 5600$$ Calculate objective at B: $$Z_B = 0.03(154838.71) + 0.04(96774.19) = 4645.16 + 3870.97 = 8516.13$$ 7. **Advice to KML Management:** Send approximately 154839 promotions to current customers and 96774 promotions to new customers to maximize sales. 8. **Excel Solution and Sensitivity Analysis:** - i. Increasing budget by 1000 will increase feasible region slightly, potentially increasing sales. - ii. Decreasing minimum current customer test drives to 3000 reduces $x \geq 120000$ to $x \geq 12000$, expanding feasible region and possibly changing optimal solution. - iii. Sensitivity ranges depend on shadow prices and allowable increases/decreases in constraints. - iv. Dual value for budget indicates how much the objective function will improve per unit increase in budget; positive dual value means increasing budget increases sales. Final decision: $$x \approx 154839, \quad y \approx 96774$$