1. **Problem 1**: Determine if there exist non-negative quantities $x_1, x_2, x_3$ of Products 1, 2, 3 such that the labor hours in departments A, B, C fully use the monthly capacities. Given hours per unit: \n Department A: $3x_1 + 2x_2 + 5x_3 = 1800$\n Department B: $4x_1 + x_2 + 3x_3 = 1450$\n Department C: $2x_1 + 4x_2 + x_3 = 1900$ Solve the system: 2. **Problem 2**: Similar to problem 1 with new data: Department A: $4x_1 + 5x_2 + 2x_3 = 1600$\n Department B: $3x_1 + 2x_2 + 3x_3 = 800$\n Department C: $x_1 + 4x_2 + 2x_3 = 1800$ 3. **Problem 3**: Find $x_1, x_2, x_3$ satisfying: \n Labor: $3x_1 + 2x_2 + 4x_3 = 1500$\n Material: $10x_1 + 8x_2 + 6x_3 = 3800$\n Units total: $x_1 + x_2 + x_3 = 500$ 4. **Problem 4**: Similar to problem 3: \n Labor: $5x_1 + 2x_2 + 4x_3 = 1300$\n Material: $15x_1 + 10x_2 + 12x_3 = 4700$\n Units total: $x_1 + x_2 + x_3 = 400$ 5. **Problem 5**: Blend 60,000 gallons with components 1,2,3. Cost and constraints: $x_1 + x_2 + x_3 = 60000$ gallons $2x_1 + 1.5x_2 + 1.25x_3 = 90000$ cost $x_1 = 2x_3$ 6. **Problem 6**: Investment $x_1, x_2, x_3$ with total $500,000$, returns $15\%, 10\%, 18\%$, average return 15\%, with $x_3 = 0.4(x_1 + x_2 + x_3)$. Solve systems for all problems: --- ### Solution approach combined: Solving each linear system: **Problem 1:** From Department C: $2x_1 +4x_2 + x_3 = 1900$ Express $x_3$: $x_3 = 1900 - 2x_1 - 4x_2$ Substitute into A and B: $3x_1 + 2x_2 + 5(1900 - 2x_1 - 4x_2) =1800$ \Rightarrow $3x_1 + 2x_2 + 9500 - 10x_1 - 20x_2=1800$ $-7x_1 - 18x_2 = -7700$ (1) $4x_1 + x_2 + 3(1900 - 2x_1 - 4x_2) =1450$ \Rightarrow $4x_1 + x_2 + 5700 - 6x_1 - 12x_2=1450$ $-2x_1 - 11x_2 = -4250$ (2) Solve (1) and (2): Multiply (2) by 3.5: $-7x_1 - 38.5x_2 = -14875$ Subtract from (1): $-7x_1 - 18x_2 - (-7x_1 - 38.5x_2) = -7700 + 14875$ $20.5x_2 = 7175 \\ x_2 = 350$ Plug $x_2=350$ into (2): $-2x_1 - 11(350) = -4250 \\ -2x_1 -3850 = -4250 \\ -2x_1 = -400 \\ x_1=200$ Then $x_3 = 1900 - 2(200) - 4(350) = 1900 - 400 - 1400 = 100$ All $x_i \\geq 0$; solution valid. **Problem 1 answer:** $x_1=200, x_2=350, x_3=100$ --- **Problem 2:** Equation system: $4x_1 + 5x_2 + 2x_3=1600$ $3x_1 + 2x_2 + 3x_3=800$ $x_1 + 4x_2 + 2x_3=1800$ From third: $x_1 = 1800 - 4x_2 - 2x_3$ Substitute into first two: First: $4(1800 -4x_2 - 2x_3) + 5x_2 + 2x_3=1600$ $7200 - 16x_2 - 8x_3 +5x_2 + 2x_3 =1600$ $7200 - 11x_2 -6x_3 =1600$ $-11x_2 -6x_3 = -5600$ (3) Second: $3(1800 - 4x_2 -2x_3) + 2x_2 + 3x_3=800$ $5400 -12x_2 -6x_3 + 2x_2 +3x_3=800$ $5400 - 10x_2 - 3x_3=800$ $-10x_2 - 3x_3 = -4600$ (4) Multiply (4) by 2: $-20x_2 -6x_3 = -9200$ Subtract (3): $(-20x_2 -6x_3) - (-11x_2 -6x_3) = -9200 + 5600$ $-9x_2 = -3600 \\ x_2=400$ Plug into (4): $-10(400) -3x_3 = -4600 \\ -4000 -3x_3 = -4600 \\ -3x_3 = -600 \\ x_3=200$ Find $x_1$: $x_1 = 1800 - 4(400) - 2(200) = 1800 -1600 -400 = -200$ Negative $x_1$ means no non-negative solution exists. **Problem 2 answer:** No non-negative solution to consume all labor. --- **Problem 3:** Solve $x_1 + x_2 + x_3=500$ $3x_1 + 2x_2 + 4x_3=1500$ $10x_1 + 8x_2 + 6x_3=3800$ Multiply first by 2: $2x_1 + 2x_2 + 2x_3 = 1000$ Subtract from second: $(3x_1 + 2x_2 + 4x_3) - (2x_1 + 2x_2 + 2x_3) = 1500 - 1000$ $x_1 + 0x_2 + 2x_3 = 500$ Multiply first by 8: $8x_1 + 8x_2 + 8x_3=4000$ Subtract from third: $(10x_1 + 8x_2 + 6x_3) - (8x_1 +8x_2 +8x_3) = 3800 - 4000$ $2x_1 + 0x_2 - 2x_3 = -200$ Solve system: $x_1 + 2x_3=500$ (A) $2x_1 - 2x_3 = -200$ (B) Multiply (A) by 2: $2x_1 + 4x_3=1000$ Subtract (B): $(2x_1 + 4x_3) - (2x_1 - 2x_3) = 1000 - (-200)$ $6x_3 = 1200 \\ x_3 = 200$ From (A): $x_1 + 2(200) = 500 \\ x_1 + 400 = 500 \\ x_1=100$ From sum: $x_1 + x_2 + x_3=500 \\ 100 + x_2 + 200 = 500 \\ x_2 = 200$ **Problem 3 answer:** $x_1=100$, $x_2=200$, $x_3=200$ --- **Problem 4:** Solve $x_1 + x_2 + x_3=400$ $5x_1 + 2x_2 + 4x_3=1300$ $15x_1 + 10x_2 + 12x_3=4700$ Multiply first by 2: $2x_1 + 2x_2 + 2x_3=800$ Subtract from second: $(5x_1 + 2x_2 + 4x_3) - (2x_1 + 2x_2 + 2x_3) = 1300 - 800$ $3x_1 + 0x_2 + 2x_3 = 500$ Multiply first by 10: $10x_1 + 10x_2 + 10x_3=4000$ Subtract from third: $(15x_1 + 10x_2 + 12x_3) - (10x_1 + 10x_2 + 10x_3) = 4700 - 4000$ $5x_1 + 0x_2 + 2x_3 = 700$ Solve system: $3x_1 + 2x_3=500$ $5x_1 + 2x_3=700$ Subtract first from second: $2x_1=200 \\ x_1=100$ Plug into first: $3(100) + 2x_3=500 \\ 300 + 2x_3=500 \\ 2x_3=200 \\ x_3=100$ From sum: $100 + x_2 + 100=400 \\ x_2=200$ **Problem 4 answer:** $x_1=100$, $x_2=200$, $x_3=100$ --- **Problem 5:** Variables $x_1, x_2, x_3$ gallons of components: $x_1 + x_2 + x_3=60000$ $2x_1 + 1.5x_2 + 1.25x_3 = 90000$ $x_1=2x_3$ Replace $x_1$: $2x_3 + x_2 + x_3=60000 \\ x_2 + 3x_3=60000$ Cost: $2(2x_3) + 1.5x_2 + 1.25x_3=90000 \\ 4x_3 + 1.5x_2 + 1.25x_3=90000 \\ 5.25x_3 + 1.5x_2=90000$ From first: $x_2=60000 - 3x_3$ Substitute: $5.25x_3 + 1.5(60000 - 3x_3) = 90000 \\ 5.25x_3 + 90000 - 4.5x_3=90000 \\ 0.75x_3=0 \\ x_3=0$ Then $x_1=2x_3=0$, $x_2=60000$ **Problem 5 answer:** $x_1=0, x_2=60000, x_3=0$ --- **Problem 6:** Investments $x_1, x_2, x_3$ with total $500,000$: $x_1 + x_2 + x_3=500000$ $0.15x_1 + 0.10x_2 + 0.18x_3 = 0.15(500000) = 75000$ $x_3 = 0.4 imes 500000 = 200000$ Substitute $x_3$: $x_1 + x_2 = 300000$ Interest: $0.15x_1 + 0.10x_2 + 0.18(200000) = 75000 \\ 0.15x_1 + 0.10x_2 = 75000 - 36000 = 39000$ From $x_1 = 300000 - x_2$: $0.15(300000 - x_2) + 0.10x_2=39000 \\ 45000 - 0.15x_2 + 0.10x_2=39000 \\ -0.05x_2 = -6000 \\ x_2=120000$ Then $x_1=180000$ **Problem 6 answer:** $x_1=180000, x_2=120000, x_3=200000$ --- **Summary:** - Problem 1: $x_1=200, x_2=350, x_3=100$ - Problem 2: No non-negative solution - Problem 3: $x_1=100, x_2=200, x_3=200$ - Problem 4: $x_1=100, x_2=200, x_3=100$ - Problem 5: $x_1=0, x_2=60000, x_3=0$ - Problem 6: $x_1=180000, x_2=120000, x_3=200000$