Let's solve the linear programming problem step by step: ### **Step 1: Write the constraints and the objective function** Objective: \( p = x - 2y \) (maximize) Subject to: 1. \( x + 2y \leq 8 \) 2. \( x - 6y \leq 0 \) (or \( x \leq 6y \)) 3. \( 3x - 2y \geq 0 \) (or \( 3x \geq 2y \) or \( y \leq \frac{3}{2}x \)) 4. \( x \geq 0, y \geq 0 \) --- ### **Step 2: Sketch the Feasible Region** #### **Turn inequalities into equations:** - \( x + 2y = 8 \) - \( x = 6y \) - \( 3x = 2y \) or \( y = \frac{3}{2}x \) #### **Find Intersection Points (Vertices):** Let's solve the constraints pairwise to find the feasible region's vertices. --- #### **A. \( x + 2y = 8 \) and \( x = 6y \)** Substitute \( x = 6y \) into the first equation: \[ 6y + 2y = 8 \implies 8y = 8 \implies y = 1, \ x = 6 \] So, intersection point: \( (6, 1) \) --- #### **B. \( x + 2y = 8 \) and \( y = \frac{3}{2}x \)** Substitute \( y = \frac{3}{2}x \) into the first equation: \[ x + 2\left(\frac{3}{2}x\right) = 8 \implies x + 3x = 8 \implies 4x = 8 \implies x = 2, \ y = 3 \] So, intersection point: \( (2, 3) \) --- #### **C. \( x = 6y \) and \( y = \frac{3}{2}x \)** Substitute \( x = 6y \) into \( y = \frac{3}{2}x \): \[ y = \frac{3}{2}(6y) = 9y \implies y - 9y = 0 \implies -8y = 0 \implies y = 0, \ x = 0 \] So, intersection point: \( (0, 0) \) --- #### **D. Check intersection with axes** - Set \( x = 0 \) in \( x + 2y = 8 \): \( 2y = 8 \implies y = 4 \) ⇒ \( (0, 4) \) - Set \( y = 0 \) in \( x + 2y = 8 \): \( x = 8 \) ⇒ \( (8, 0) \) - Set \( x = 0 \) in \( x = 6y \): \( 0 = 6y \implies y = 0 \) ⇒ \( (0, 0) \) - Set \( x = 0 \) in \( y = \frac{3}{2}x \): \( y = 0 \) ⇒ \( (0, 0) \) --- ### **Step 3: Check which points are in the feasible region** Feasible region is bounded by: - \( x + 2y \leq 8 \) - \( x \leq 6y \) - \( y \leq \frac{3}{2}x \) - \( x \geq 0, y \geq 0 \) Let's test the points \( (0, 0), (0, 4), (2, 3), (6, 1), (8, 0) \): #### (0,0) - \( x + 2y = 0 \leq 8 \) - \( x - 6y = 0 \leq 0 \) - \( 3x - 2y = 0 \geq 0 \) - Yes (feasible) #### (0, 4) - \( 0 + 2(4) = 8 \leq 8 \) - \( 0 - 6(4) = -24 \leq 0 \) - \( 3(0) - 2(4) = -8 \not\geq 0 \) - **Not feasible** #### (2, 3) - \( 2 + 2(3) = 8 \leq 8 \) - \( 2 - 6(3) = 2 - 18 = -16 \leq 0 \) - \( 3(2) - 2(3) = 6 - 6 = 0 \geq 0 \) - Yes (feasible) #### (6, 1) - \( 6 + 2(1) = 8 \leq 8 \) - \( 6 - 6(1) = 0 \leq 0 \) - \( 3(6) - 2(1) = 18 - 2 = 16 \geq 0 \) - Yes (feasible) #### (8, 0) - \( 8 + 2(0) = 8 \leq 8 \) - \( 8 - 6(0) = 8 \not\leq 0 \) - Not feasible --- ### **Step 4: Evaluate the Objective Function at Feasible Vertices** Feasible points: \( (0, 0), (2, 3), (6, 1) \) Calculate \( p = x - 2y \): - At \( (0, 0) \): \( p = 0 - 2(0) = 0 \) - At \( (2, 3) \): \( p = 2 - 2(3) = 2 - 6 = -4 \) - At \( (6, 1) \): \( p = 6 - 2(1) = 6 - 2 = 4 \) --- ### **Step 5: Conclusion** #### **Maximum value is at \( (6, 1) \) with \( p = 4 \).** --- ### **Summary Table** | Point | \( p = x - 2y \) | |----------|------------------| | (0, 0) | 0 | | (2, 3) | -4 | | (6, 1) | **4** | --- ## **Final Answer** The maximum value of \( p = x - 2y \) is **4** at the point **(6, 1)**. --- **(If you need the sketch of the feasible region, let me know! I can describe or draw it for you.)**