Let's solve the linear programming problem step by step: ### **Step 1: Write the Constraints** The constraints are: 1. \( x + 2y \leq 8 \) 2. \( x - 6y \leq 0 \) 3. \( 3x - 2y \geq 0 \) 4. \( x \geq 0 \) 5. \( y \geq 0 \) ### **Step 2: Convert Inequalities to Equalities for Graphing** 1. \( x + 2y = 8 \) &rarr; (Line 1) 2. \( x - 6y = 0 \) &rarr; (Line 2) 3. \( 3x - 2y = 0 \) &rarr; (Line 3) 4. \( x = 0 \) &rarr; (y-axis) 5. \( y = 0 \) &rarr; (x-axis) ### **Step 3: Find Intersection Points (Vertices)** Let's solve pairs of equations to get the vertices: #### **A. Intersection of (1) and (2):** \[ x + 2y = 8 \] \[ x - 6y = 0 \implies x = 6y \] Substitute \( x = 6y \) into (1): \[ 6y + 2y = 8 \implies 8y = 8 \implies y = 1 \] \[ x = 6y = 6 \times 1 = 6 \] **Vertex 1:** \( (6, 1) \) --- #### **B. Intersection of (1) and (3):** \[ x + 2y = 8 \] \[ 3x - 2y = 0 \implies 3x = 2y \implies x = \frac{2}{3}y \] Substitute \( x = \frac{2}{3}y \) into (1): \[ \frac{2}{3}y + 2y = 8 \implies \frac{2y + 6y}{3} = 8 \implies \frac{8y}{3} = 8 \implies 8y = 24 \implies y = 3 \] \[ x = \frac{2}{3} \times 3 = 2 \] **Vertex 2:** \( (2, 3) \) --- #### **C. Intersection of (2) and (3):** \[ x - 6y = 0 \implies x = 6y \] \[ 3x - 2y = 0 \implies 3x = 2y \implies x = \frac{2}{3}y \] Set \( 6y = \frac{2}{3}y \): \[ 6y = \frac{2}{3}y \implies 18y = 2y \implies 16y = 0 \implies y = 0 \] \[ x = 6y = 0 \] **Vertex 3:** \( (0, 0) \) --- #### **D. Intersection with axes** - For \( x + 2y = 8 \) and \( x = 0 \): \( 0 + 2y = 8 \implies y = 4 \) &rarr; \( (0, 4) \) - For \( x + 2y = 8 \) and \( y = 0 \): \( x = 8 \) &rarr; \( (8, 0) \) - For \( x - 6y = 0 \) and \( y = 0 \): \( x = 0 \) &rarr; \( (0, 0) \) - For \( 3x - 2y = 0 \) and \( y = 0 \): \( x = 0 \) &rarr; \( (0, 0) \) ### **Step 4: Test Feasibility of Vertices** Check which of these points satisfy all original inequalities: | Point | \(x + 2y \leq 8\) | \(x - 6y \leq 0\) | \(3x - 2y \geq 0\) | \(x \geq 0\) | \(y \geq 0\) | Feasible? | |------------|-------------------|-------------------|--------------------|--------------|--------------|-----------| | (6, 1) | 6+2=8 ≤ 8 | 6-6=0 ≤ 0 | 18-2=16 ≥ 0 | Yes | Yes | Yes | | (2, 3) | 2+6=8 ≤ 8 | 2-18=-16 ≤ 0 | 6-6=0 ≥ 0 | Yes | Yes | Yes | | (0, 0) | 0+0=0 ≤ 8 | 0-0=0 ≤ 0 | 0-0=0 ≥ 0 | Yes | Yes | Yes | | (0, 4) | 0+8=8 ≤ 8 | 0-24=-24 ≤ 0 | 0-8=-8 < 0 | Yes | Yes | **No** | | (8, 0) | 8+0=8 ≤ 8 | 8-0=8 > 0 | 24-0=24 ≥ 0 | Yes | Yes | **No** | So, feasible vertices are: **(0, 0), (2, 3), (6, 1)** --- ### **Step 5: Evaluate Objective Function at Each Vertex** \( p = x - 2y \) - At \( (0,0) \): \( p = 0 - 2 \times 0 = 0 \) - At \( (2,3) \): \( p = 2 - 2 \times 3 = 2 - 6 = -4 \) - At \( (6,1) \): \( p = 6 - 2 \times 1 = 6 - 2 = 4 \) ### **Step 6: Final Answer** **Maximum value of \( p \) is \(\boxed{4}\) at the point \((6, 1)\).** --- ### **Step 7: Sketch the Feasible Region (optional description)** - Plot the lines \( x + 2y = 8 \), \( x - 6y = 0 \), \( 3x - 2y = 0 \), and the axes \( x=0 \), \( y=0 \). - The feasible region is the area that satisfies all inequalities. - Mark the feasible vertices \( (0,0) \), \( (2,3) \), \( (6,1) \). - The maximum occurs at \( (6,1) \). --- **If you want a graph, let me know!**