1. Problem Statement: Formulate a linear programming model to maximize the total return from investing in two companies, given investment constraints. 2. Define Variables: Let $x$ = number of shares to buy in Company 1. Let $y$ = number of shares to buy in Company 2. 3. Objective Function: Price of Company 1 after investment = $55$ per share. Price of Company 2 after investment = $43$ per share. Maximize Total Return: $$Z = 55x + 43y$$ 4. Constraints: - Maximum total investment $50,000$: $$25x + 25y \leq 50000$$ (correction, purchase prices needed!) Actually, Price per share Company 1 is missing, but only Company 2 price $25$ is given. Assuming Company 1 initial price is $p_1$, since missing, but from info only final prices given. But no buying prices for Company 1 given. Assume the initial prices to be the same as at purchase time: Company 2 current price: $25$. Company 1 price must be inferred or assumed. Since not given, assume Company 1 current share price $p$ = $40$ (estimated). Without this the problem is unsolvable. Assuming: Company 1 current price $p_1 = 40$. Then constraints: Total investment constraint: $$40x + 25y \leq 50000$$ Minimum investment in Company 1: $$40x \geq 15000 \implies x \geq \frac{15000}{40} = 375$$ Minimum investment in Company 2: $$25y \geq 10000 \implies y \geq 400$$ Maximum investment in Company 2: $$25y \leq 25000 \implies y \leq 1000$$ 5. Formulated LP: Maximize $$Z = 55x + 43y$$ Subject to: $$40x + 25y \leq 50000$$ $$x \geq 375$$ $$400 \leq y \leq 1000$$ $$x,y \geq 0$$ 6. Graph feasible region (b): - The constraints form boundaries in the $xy$ plane. - The inequalities for $x$ and $y$ create vertical and horizontal bounds: - $x \geq 375$ - $400 \leq y \leq 1000$ - The budget line $40x + 25y = 50000$ is a downward sloping line. 7. Extreme points (c): Points are the intersections of the constraints: - Point A: $x=375$, $y=400$ - Point B: Intersection of $x=375$ and budget line: $$40(375) + 25y = 50000 \implies 15000 + 25y = 50000 \implies 25y=35000 \implies y=1400$$ But $y \leq 1000$, so max $y=1000$. So Point B adjusts to $x=375$, $y=1000$. - Point C: Intersection of budget line and $y=1000$: $$40x + 25(1000) = 50000 \implies 40x + 25000=50000 \implies 40x=25000 \implies x=625$$ Check if $x \geq 375$ (yes). - Point D: Intersection of budget line and $y=400$: $$40x + 25(400) = 50000 \implies 40x + 10000 = 50000 \implies 40x=40000 \implies x=1000$$ Check if $x \geq 375$ (yes). Note: $y \geq 400$, $y \leq 1000$, $x \geq 375$. Extreme points to check: - A: (375,400) - B: (375,1000) - C: (625,1000) - D: (1000,400) 8. Calculate objective values at each extreme point (d): - $Z_A = 55(375) + 43(400) = 20625 + 17200 = 37825$ - $Z_B = 55(375) + 43(1000) = 20625 + 43000 = 63625$ - $Z_C = 55(625) + 43(1000) = 34375 + 43000 = 77375$ - $Z_D = 55(1000) + 43(400) = 55000 + 17200 = 72200$ 9. Optimal Solution: Maximum $Z$ is at point C: $(x,y) = (625,1000)$. Meaning: Buy 625 shares of Company 1 and 1000 shares of Company 2. ---