1. **Problem 1: 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. **Problem 2: Minimize $80x + 60y$ subject to constraints:** $$2x + 6y \geq 18$$ $$6x + 3y \leq 24$$ $$x + y \geq 4$$ $$x \geq 0, y \geq 0$$ --- ### Step-by-step solution for Problem 1: 1. **Identify feasible region:** - From $3x + 4y \geq 12$, rewrite as $y \geq \frac{12 - 3x}{4}$. - From $4x + 2y \leq 20$, rewrite as $y \leq \frac{20 - 4x}{2} = 10 - 2x$. - $x \leq 4$, $y \geq 6$, $x \geq 0$, $y \geq 0$. 2. **Find intersection points of boundary lines that form the feasible polygon:** - Intersection of $3x + 4y = 12$ and $4x + 2y = 20$: Solve system: $$3x + 4y = 12$$ $$4x + 2y = 20$$ Multiply second by 2: $$8x + 4y = 40$$ Subtract first from this: $$8x + 4y - (3x + 4y) = 40 - 12 \Rightarrow 5x = 28 \Rightarrow x = \frac{28}{5} = 5.6$$ Substitute back: $$3(5.6) + 4y = 12 \Rightarrow 16.8 + 4y = 12 \Rightarrow 4y = -4.8 \Rightarrow y = -1.2$$ Since $y = -1.2$ is not feasible ($y \geq 0$), discard this point. - Intersection of $3x + 4y = 12$ and $y = 6$: $$3x + 4(6) = 12 \Rightarrow 3x + 24 = 12 \Rightarrow 3x = -12 \Rightarrow x = -4$$ Not feasible since $x \geq 0$. - Intersection of $4x + 2y = 20$ and $y = 6$: $$4x + 2(6) = 20 \Rightarrow 4x + 12 = 20 \Rightarrow 4x = 8 \Rightarrow x = 2$$ Point $(2,6)$ is feasible. - Intersection of $x = 4$ and $y = 6$: Point $(4,6)$ feasible. - Intersection of $x = 4$ and $4x + 2y = 20$: $$4(4) + 2y = 20 \Rightarrow 16 + 2y = 20 \Rightarrow 2y = 4 \Rightarrow y = 2$$ Point $(4,2)$ but $y \geq 6$ not satisfied, discard. - Intersection of $x = 4$ and $3x + 4y = 12$: $$3(4) + 4y = 12 \Rightarrow 12 + 4y = 12 \Rightarrow 4y = 0 \Rightarrow y = 0$$ Not feasible since $y \geq 6$. - Check corners from constraints: - $(0,6)$ satisfies all constraints. - $(0, \text{from } 3x + 4y = 12)$: $3(0) + 4y = 12 \Rightarrow y = 3$ not feasible since $y \geq 6$. 3. **Feasible region vertices are:** - $(0,6)$ - $(2,6)$ - $(4,6)$ 4. **Evaluate objective $30x + 50y$ at vertices:** - 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. **Maximum value is $420$ at $(4,6)$**. --- ### Step-by-step solution for Problem 2: 1. **Identify feasible region:** - From $2x + 6y \geq 18$, rewrite as $y \geq \frac{18 - 2x}{6} = 3 - \frac{x}{3}$. - From $6x + 3y \leq 24$, rewrite as $y \leq \frac{24 - 6x}{3} = 8 - 2x$. - From $x + y \geq 4$, rewrite as $y \geq 4 - x$. - $x \geq 0$, $y \geq 0$. 2. **Find intersection points of boundary lines:** - Intersection of $2x + 6y = 18$ and $6x + 3y = 24$: Multiply first by 1 and second by 2: $$2x + 6y = 18$$ $$12x + 6y = 48$$ Subtract first from second: $$10x = 30 \Rightarrow x = 3$$ Substitute back: $$2(3) + 6y = 18 \Rightarrow 6 + 6y = 18 \Rightarrow 6y = 12 \Rightarrow y = 2$$ Point $(3,2)$. - Intersection of $2x + 6y = 18$ and $x + y = 4$: From $x + y = 4$, $y = 4 - x$. Substitute into first: $$2x + 6(4 - x) = 18 \Rightarrow 2x + 24 - 6x = 18 \Rightarrow -4x = -6 \Rightarrow x = 1.5$$ Then $y = 4 - 1.5 = 2.5$. Point $(1.5, 2.5)$. - Intersection of $6x + 3y = 24$ and $x + y = 4$: From $x + y = 4$, $y = 4 - x$. Substitute: $$6x + 3(4 - x) = 24 \Rightarrow 6x + 12 - 3x = 24 \Rightarrow 3x = 12 \Rightarrow x = 4$$ Then $y = 4 - 4 = 0$. Point $(4,0)$. - Check non-negativity and constraints for these points: All points satisfy $x \geq 0$, $y \geq 0$. 3. **Feasible region vertices are:** - $(1.5, 2.5)$ - $(3, 2)$ - $(4, 0)$ 4. **Evaluate objective $80x + 60y$ at vertices:** - At $(1.5, 2.5)$: $80(1.5) + 60(2.5) = 120 + 150 = 270$ - At $(3, 2)$: $80(3) + 60(2) = 240 + 120 = 360$ - At $(4, 0)$: $80(4) + 60(0) = 320 + 0 = 320$ 5. **Minimum value is $270$ at $(1.5, 2.5)$**. --- ### Final answers: - Problem 1 maximum: $420$ at $(4,6)$. - Problem 2 minimum: $270$ at $(1.5, 2.5)$.