To solve this problem and find the minimum value of $z = 5x + 3y$ subject to the constraints, we will follow the linear programming approach. The steps include:

1. Write down the constraints:

$6x + 2y \geq 120, \quad 8x + 6y \geq 240, \quad 2x + 4y \geq 80, \quad x \geq 0, \quad y \geq 0$

Rewriting them in slope-intercept form:

1. $y \geq 60 - 3x$ from $6x + 2y = 120$,
2. $y \geq 40 - \frac{4}{3}x$ from $8x + 6y = 240$,
3. $y \geq 20 - \frac{1}{2}x$ from $2x + 4y = 80$.

2. Identify the feasible region:

The feasible region is bounded by the intersection of these inequalities. Since $x \geq 0$ and $y \geq 0$, we consider only the first quadrant.

3. Find the corner points:

We solve the constraints pairwise to find intersection points. Using elimination or substitution:

• Intersection of $6x + 2y = 120$ and $8x + 6y = 240$:

&6x + 2y = 120 \quad \text{(Multiply by 3:)} \quad 18x + 6y = 360, \\ &8x + 6y = 240 \quad \text{(Subtract equations:)} \quad 10x = 120 \implies x = 12, \quad y = 60 - 3(12) = 24.$$ - Intersection of $$6x + 2y = 120$$ and $$2x + 4y = 80$$: $$\begin{aligned} &6x + 2y = 120 \quad \text{(Divide by 2:)} \quad 3x + y = 60, \\ &2x + 4y = 80 \quad \text{(Divide by 2:)} \quad x + 2y = 40 \quad \text{(Solve simultaneously).}$$ Multiply $$3x + y = 60$$ by 2: $$6x + 2y = 120 \quad \text{and subtract:} \quad x = 0, \quad y = 40.$$ - Intersection of $$8x + 6y = 240$$ and $$2x + 4y = 80$$: $$\begin{aligned} &8x + 6y = 240 \quad \text{(Divide by 2:)} \quad 4x + 3y = 120, \\ &2x + 4y = 80 \quad \text{(Multiply by 3:)} \quad 6x + 12y = 240 \quad \text{(Solve simultaneously).}$$ Multiply $$4x + 3y = 120$$ by 4: $$16x + 12y = 480 \quad \text{and subtract:} \quad 10x = 240 \implies x = 24, \quad y = 40 - \frac{1}{2}(24) = 28.$$ ### 4. Evaluate $$z = 5x + 3y$$ at the corner points: $$\text{Corner Points: } (12, 24), (0, 40), (24, 28).$$ 1. At $$(12, 24)$$: $$z = 5(12) + 3(24) = 60 + 72 = 132.$$ 2. At $$(0, 40)$$: $$z = 5(0) + 3(40) = 0 + 120 = 120.$$ 3. At $$(24, 28)$$: $$z = 5(24) + 3(28) = 120 + 84 = 204.$$ ### 5. Conclusion: The minimum value of $$z = 120$$, occurring at $$(x, y) = (0, 40)$$.