1. **State the problem:** Maximize $30x + 50y$ subject to constraints: $$3x + 4y \geq 12$$ $$4x + 2y \leq 20$$ $$x \leq 4$$ $$y \geq 6$$ $$x \geq 0$$ $$y \geq 0$$ 2. **Find intersection points of constraint lines to identify vertices of the feasible region:** - Intersection of $3x + 4y = 12$ and $4x + 2y = 20$: Multiply second equation by 2: $$8x + 4y = 40$$ Subtract first equation: $$(8x + 4y) - (3x + 4y) = 40 - 12$$ $$5x = 28 \Rightarrow x = \frac{28}{5} = 5.6$$ Substitute back into $3x + 4y = 12$: $$3(5.6) + 4y = 12$$ $$16.8 + 4y = 12$$ $$4y = 12 - 16.8 = -4.8$$ $$y = -1.2$$ (Not feasible since $y \geq 6$) - Intersection of $3x + 4y = 12$ and $y = 6$: $$3x + 4(6) = 12$$ $$3x + 24 = 12$$ $$3x = -12$$ $$x = -4$$ (Not feasible since $x \geq 0$) - Intersection of $4x + 2y = 20$ and $y = 6$: $$4x + 2(6) = 20$$ $$4x + 12 = 20$$ $$4x = 8$$ $$x = 2$$ (Feasible since $x \leq 4$, $x \geq 0$, $y \geq 6$) - Intersection of $x = 4$ and $4x + 2y = 20$: $$4(4) + 2y = 20$$ $$16 + 2y = 20$$ $$2y = 4$$ $$y = 2$$ (Not feasible since $y \geq 6$) - Intersection of $x = 4$ and $y = 6$: Point $(4,6)$ (Feasible) - Intersection of $x = 4$ and $3x + 4y = 12$: $$3(4) + 4y = 12$$ $$12 + 4y = 12$$ $$4y = 0$$ $$y = 0$$ (Not feasible since $y \geq 6$) - Intersection of $x = 0$ and $3x + 4y = 12$: $$3(0) + 4y = 12$$ $$4y = 12$$ $$y = 3$$ (Not feasible since $y \geq 6$) - Intersection of $x = 0$ and $y = 6$: Point $(0,6)$ (Feasible) 3. **Feasible region vertices:** Points satisfying all constraints are: - $(0,6)$ - $(2,6)$ - $(4,6)$ 4. **Evaluate objective function $30x + 50y$ at each vertex:** - At $(0,6)$: $30(0) + 50(6) = 300$ - At $(2,6)$: $30(2) + 50(6) = 60 + 300 = 360$ - At $(4,6)$: $30(4) + 50(6) = 120 + 300 = 420$ 5. **Conclusion:** Maximum value is $420$ at point $(4,6)$. --- 1. **State the problem:** Minimize $80x + 60y$ subject to constraints: $$2x + 6y \geq 18$$ $$6x + 3y \leq 24$$ $$x + y \geq 4$$ $$x \geq 0$$ $$y \geq 0$$ 2. **Find intersection points of constraint lines:** - Intersection of $2x + 6y = 18$ and $6x + 3y = 24$: Multiply first by 3: $$6x + 18y = 54$$ Subtract second: $$(6x + 18y) - (6x + 3y) = 54 - 24$$ $$15y = 30 \Rightarrow y = 2$$ Substitute into $2x + 6y = 18$: $$2x + 6(2) = 18$$ $$2x + 12 = 18$$ $$2x = 6$$ $$x = 3$$ - Intersection of $2x + 6y = 18$ and $x + y = 4$: From $x + y = 4$, $x = 4 - y$ Substitute into $2x + 6y = 18$: $$2(4 - y) + 6y = 18$$ $$8 - 2y + 6y = 18$$ $$8 + 4y = 18$$ $$4y = 10$$ $$y = 2.5$$ $$x = 4 - 2.5 = 1.5$$ - Intersection of $6x + 3y = 24$ and $x + y = 4$: From $x + y = 4$, $y = 4 - x$ Substitute into $6x + 3y = 24$: $$6x + 3(4 - x) = 24$$ $$6x + 12 - 3x = 24$$ $$3x + 12 = 24$$ $$3x = 12$$ $$x = 4$$ $$y = 4 - 4 = 0$$ 3. **Check feasibility of points (must satisfy all inequalities):** - $(3,2)$: $$2(3) + 6(2) = 6 + 12 = 18 \geq 18 \checkmark$$ $$6(3) + 3(2) = 18 + 6 = 24 \leq 24 \checkmark$$ $$3 + 2 = 5 \geq 4 \checkmark$$ - $(1.5, 2.5)$: $$2(1.5) + 6(2.5) = 3 + 15 = 18 \geq 18 \checkmark$$ $$6(1.5) + 3(2.5) = 9 + 7.5 = 16.5 \leq 24 \checkmark$$ $$1.5 + 2.5 = 4 \geq 4 \checkmark$$ - $(4,0)$: $$2(4) + 6(0) = 8 \geq 18 \times$$ (Fails) - $(0,3)$ (from $2x + 6y = 18$ with $x=0$): $$0 + 6(3) = 18 \geq 18 \checkmark$$ $$6(0) + 3(3) = 9 \leq 24 \checkmark$$ $$0 + 3 = 3 \geq 4 \times$$ (Fails) - $(0,4)$ (from $x + y = 4$): $$2(0) + 6(4) = 24 \geq 18 \checkmark$$ $$6(0) + 3(4) = 12 \leq 24 \checkmark$$ $$0 + 4 = 4 \geq 4 \checkmark$$ 4. **Feasible vertices:** - $(3,2)$ - $(1.5, 2.5)$ - $(0,4)$ 5. **Evaluate objective function $80x + 60y$ at each vertex:** - At $(3,2)$: $80(3) + 60(2) = 240 + 120 = 360$ - At $(1.5, 2.5)$: $80(1.5) + 60(2.5) = 120 + 150 = 270$ - At $(0,4)$: $80(0) + 60(4) = 0 + 240 = 240$ 6. **Conclusion:** Minimum value is $240$ at point $(0,4)$.