1. **State the problem:** We want to determine how many wrenches ($x$) and pliers ($y$) to produce to maximize profit given constraints on steel, machine hours, and demand. 2. **Define variables:** Let $x$ = number of wrenches produced Let $y$ = number of pliers produced 3. **Write the objective function:** Maximize profit: $$P = 0.40x + 0.30y$$ 4. **Write the constraints from the table:** - Steel constraint: $$1.5x + 1.0y \leq 15000$$ - Molding machine hours: $$1.0x + 1.0y \leq 12000$$ - Assembly machine hours: $$0.4x + 0.5y \leq 5000$$ - Demand limits: $$x \leq 8000$$ $$y \leq 10000$$ - Non-negativity: $$x \geq 0, \quad y \geq 0$$ 5. **Explain the approach:** This is a linear programming problem. We find the feasible region defined by the constraints and evaluate the objective function at the vertices to find the maximum profit. 6. **Graph the constraints:** - Steel: $y \leq \frac{15000 - 1.5x}{1}$ - Molding: $y \leq 12000 - x$ - Assembly: $y \leq \frac{5000 - 0.4x}{0.5} = 10000 - 0.8x$ - Demand: $x \leq 8000$, $y \leq 10000$ 7. **Find intersection points (vertices) of the feasible region:** - Intersection of steel and molding: $$1.5x + y = 15000$$ $$x + y = 12000$$ Subtract second from first: $$0.5x = 3000 \Rightarrow x = 6000$$ Then $y = 12000 - 6000 = 6000$ - Intersection of steel and assembly: $$1.5x + y = 15000$$ $$0.4x + 0.5y = 5000$$ Multiply second by 2: $$0.8x + y = 10000$$ Subtract first from this: $$0.8x + y - (1.5x + y) = 10000 - 15000$$ $$-0.7x = -5000 \Rightarrow x = \frac{5000}{0.7} \approx 7142.86$$ Then $y = 15000 - 1.5(7142.86) = 15000 - 10714.29 = 4285.71$ - Intersection of molding and assembly: $$x + y = 12000$$ $$0.4x + 0.5y = 5000$$ Multiply first by 0.5: $$0.5x + 0.5y = 6000$$ Subtract second: $$0.5x + 0.5y - (0.4x + 0.5y) = 6000 - 5000$$ $$0.1x = 1000 \Rightarrow x = 10000$$ Then $y = 12000 - 10000 = 2000$ 8. **Check demand limits:** All $x$ and $y$ values are within demand limits. 9. **Evaluate profit at vertices:** - At (0,0): $P=0$ - At (0,10000): $P=0.40(0)+0.30(10000)=3000$ - At (8000,0): $P=0.40(8000)+0=3200$ - At (6000,6000): $P=0.40(6000)+0.30(6000)=2400+1800=4200$ - At (7142.86,4285.71): $P=0.40(7142.86)+0.30(4285.71)=2857.14+1285.71=4142.85$ - At (10000,2000) violates $x \leq 8000$ demand limit, so discard. 10. **Conclusion:** Maximum profit is approximately $4200$ at $x=6000$ wrenches and $y=6000$ pliers.