1. **Problem Statement:** We want to maximize profit from manufacturing components A, B, and C given labor-hour constraints. 2. **Define Variables:** Let $x$, $y$, and $z$ be the number of components A, B, and C produced respectively. 3. **Objective Function:** The profit on each component is $7$ for A, $8$ for B, and $10$ for C. So, maximize: $$\text{Maximize } P = 7x + 8y + 10z$$ 4. **Constraints:** Given fabrication and assembly hours: - Fabrication hours: Component A requires 2 hours, B requires 3 hours, C requires 2 hours, with a max of 1000 hours: $$2x + 3y + 2z \leq 1000$$ - Assembly hours: Component A requires 1 hour, B requires 1 hour, C requires 2 hours, max 800 hours: $$x + y + 2z \leq 800$$ - Non-negativity: $$x \geq 0, y \geq 0, z \geq 0$$ 5. **Graphical Method:** Since graphical methods apply only to two variables, we can solve by fixing one variable (e.g., $z=0$) and then graph to find maximal profit. For simplicity, set $z=0$: - Constraints reduce to: $$2x + 3y \leq 1000$$ $$x + y \leq 800$$ - Objective: maximize $7x + 8y$ 6. **Simplex Method:** Set up the linear program with slack variables $s_1$, $s_2$: $$\begin{cases} 2x + 3y + 2z + s_1 = 1000\\ x + y + 2z + s_2 = 800\\ x,y,z,s_1,s_2 \geq 0 \end{cases}$$ 7. **Final Formulated LP:** $$\text{Maximize } 7x + 8y + 10z$$ subject to $$2x + 3y + 2z \leq 1000$$ $$x + y + 2z \leq 800$$ $$x,y,z \geq 0$$ The graphical method can be done for two variables cases by fixing the third. Simplex method proceeds by setting up the tableau and iteratively improving the solution until optimality is reached.