1. **State the problem:** We want to determine the number of 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 data:** - 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: $x \geq 0$, $y \geq 0$ 5. **Explain the approach:** This is a linear programming problem. The feasible region is defined by the constraints above. The optimal solution lies at a vertex of this region. 6. **Check corner points (vertices) of the feasible region:** We find intersections of constraints to evaluate $P$. - Intersection of steel and molding machine constraints: $$\begin{cases} 1.5x + y = 15000 \\ x + y = 12000 \end{cases}$$ Subtract second from first: $$1.5x + y - (x + y) = 15000 - 12000 \Rightarrow 0.5x = 3000 \Rightarrow x = 6000$$ Then $y = 12000 - 6000 = 6000$ Check assembly machine: $$0.4(6000) + 0.5(6000) = 2400 + 3000 = 5400 > 5000$$ So this point is not feasible. - Intersection of steel and assembly machine constraints: $$\begin{cases} 1.5x + y = 15000 \\ 0.4x + 0.5y = 5000 \end{cases}$$ Multiply second by 2: $$0.8x + y = 10000$$ Subtract first from this: $$(0.8x + y) - (1.5x + y) = 10000 - 15000 \Rightarrow -0.7x = -5000 \Rightarrow x = \frac{5000}{0.7} \approx 7142.86$$ Then from first: $$1.5(7142.86) + y = 15000 \Rightarrow 10714.29 + y = 15000 \Rightarrow y = 4285.71$$ Check molding machine: $$7142.86 + 4285.71 = 11428.57 \leq 12000$$ feasible. - Intersection of molding and assembly machine constraints: $$\begin{cases} x + y = 12000 \\ 0.4x + 0.5y = 5000 \end{cases}$$ From first: $y = 12000 - x$ Substitute into second: $$0.4x + 0.5(12000 - x) = 5000 \Rightarrow 0.4x + 6000 - 0.5x = 5000 \Rightarrow -0.1x = -1000 \Rightarrow x = 10000$$ Then $y = 12000 - 10000 = 2000$ Check steel: $$1.5(10000) + 2000 = 15000 + 2000 = 17000 > 15000$$ not feasible. - Check demand limits: $x \leq 8000$, $y \leq 10000$ 7. **Evaluate profit at feasible vertices:** - At $(7142.86, 4285.71)$: $$P = 0.40(7142.86) + 0.30(4285.71) = 2857.14 + 1285.71 = 4142.85$$ - At $(8000, 0)$ (demand limit for wrenches): Check constraints: Steel: $1.5(8000) + 0 = 12000 \leq 15000$ ok Molding: $8000 + 0 = 8000 \leq 12000$ ok Assembly: $0.4(8000) + 0 = 3200 \leq 5000$ ok Profit: $$0.40(8000) + 0 = 3200$$ - At $(0, 10000)$ (demand limit for pliers): Steel: $0 + 1(10000) = 10000 \leq 15000$ ok Molding: $0 + 10000 = 10000 \leq 12000$ ok Assembly: $0 + 0.5(10000) = 5000 \leq 5000$ ok Profit: $$0 + 0.30(10000) = 3000$$ 8. **Conclusion:** The maximum profit is approximately $4142.85$ at producing about $7143$ wrenches and $4286$ pliers. **Final answer:** Produce approximately $7143$ wrenches and $4286$ pliers to maximize profit of about $4142.85$.