1. The problem is to minimize the objective function $$z = 40x + 50y$$ subject to the constraints: $$2x + y \geq 24$$ $$x + y \geq 16$$ $$x + 2y \geq 20$$ $$x, y \geq 0$$ 2. We find the feasible region defined by these inequalities and then check the values of $$z$$ at each vertex (corner point) of this region. 3. Find the vertices by solving the system for the intersections of the constraints: - Intersection of $$2x + y = 24$$ and $$x + y = 16$$: Subtract equations: $$ (2x + y) - (x + y) = 24 - 16 \implies x = 8 $$ Substitute $$x=8$$ into $$x + y =16$$: $$8 + y = 16 \implies y = 8$$ So, vertex at $$(8,8)$$. - Intersection of $$2x + y = 24$$ and $$x + 2y = 20$$: From $$2x + y = 24$$, rewrite $$y = 24 - 2x$$ Substitute into $$x + 2y = 20$$: $$x + 2(24 - 2x) = 20 \implies x + 48 - 4x = 20 \implies -3x = -28 \implies x = \frac{28}{3}$$ Then $$y = 24 - 2 \times \frac{28}{3} = 24 - \frac{56}{3} = \frac{72}{3} - \frac{56}{3} = \frac{16}{3}$$ So, vertex at $$\left(\frac{28}{3}, \frac{16}{3}\right)$$. - Intersection of $$x + y = 16$$ and $$x + 2y = 20$$: From $$x + y =16$$, rewrite $$x = 16 - y$$ Substitute into $$x + 2y = 20$$: $$16 - y + 2y = 20 \implies 16 + y = 20 \implies y = 4$$ Then $$x = 16 - 4 = 12$$ So, vertex at $$(12, 4)$$. 4. Verify these vertices satisfy $$x, y \geq 0$$ and all constraints (which they do). 5. Compute $$z = 40x + 50y$$ at each vertex: - At $$(8,8)$$: $$z = 40 \times 8 + 50 \times 8 = 320 + 400 = 720$$ - At $$\left(\frac{28}{3}, \frac{16}{3}\right)$$: $$z = 40 \times \frac{28}{3} + 50 \times \frac{16}{3} = \frac{1120}{3} + \frac{800}{3} = \frac{1920}{3} = 640$$ - At $$(12,4)$$: $$z = 40 \times 12 + 50 \times 4 = 480 + 200 = 680$$ 6. The minimum value of $$z$$ among these is $$640$$ at $$\left(\frac{28}{3}, \frac{16}{3}\right)$$. **Final answer:** Minimum value of $$z$$ is $$640$$ at $$\left(\frac{28}{3}, \frac{16}{3}\right)$$.