1. **State the problem:** We want to maximize the contribution from products I and II given machine hour constraints. 2. **Define variables:** Let $x$ = number of units of product I produced. Let $y$ = number of units of product II produced. 3. **Formulate constraints based on machine hours:** Machine A is available for 100 hours: $$4x + 5y \leq 100$$ Machine B is available for 80 hours: $$5x + 2y \leq 80$$ Also, $x, y \geq 0$ (cannot produce negative units). 4. **Objective function (maximize contribution):** $$\text{Maximize } Z = 10x + 15y$$ 5. **Convert inequalities to equations with slack variables:** $$4x + 5y + s_1 = 100$$ $$5x + 2y + s_2 = 80$$ where $s_1, s_2 \geq 0$ are slack variables. 6. **Apply simplex method:** - Start with basic feasible solution $x=0, y=0, s_1=100, s_2=80$. - Compute ratios to find pivot and update variables (details omitted for brevity). 7. **Solve using substitution or graphical method:** From constraint 1: $$y = \frac{100 - 4x}{5}$$ From constraint 2: $$y = \frac{80 - 5x}{2}$$ Set equal to find intersection: $$\frac{100 - 4x}{5} = \frac{80 - 5x}{2}$$ Cross multiply: $$2(100 - 4x) = 5(80 - 5x)$$ $$200 - 8x = 400 - 25x$$ $$17x = 200$$ $$x = \frac{200}{17} \approx 11.76$$ Substitute back for $y$: $$y = \frac{100 - 4(11.76)}{5} = \frac{100 - 47.05}{5} = \frac{52.95}{5} = 10.59$$ 8. **Calculate maximum contribution:** $$Z = 10(11.76) + 15(10.59) = 117.6 + 158.85 = 276.45$$ 9. **Final answer:** The maximum contribution is approximately **276.45** when producing about **11.76** units of product I and **10.59** units of product II.