Sure! Let's solve the linear programming problem **step by step with explanations**. --- ## 1. **Write Down the Constraints** From the image, the system of inequalities is: 1. \( 6x - 2y \leq 36 \) 2. \( -x + 8y \leq 40 \) 3. \( x \geq 0 \) 4. \( y \geq 0 \) --- ## 2. **Find the Corner Points (Vertices) of the Feasible Region** These points occur where the boundary lines intersect. We'll solve for intersections by setting the inequalities as equalities. ### Step 1: **Find Intersection with the Axes** #### (a) *x-axis* (\( y = 0 \)): - \( 6x \leq 36 \implies x \leq 6 \) - \( -x \leq 40 \implies x \geq -40 \) (but \( x \geq 0 \) by constraint) So, possible points: **(0, 0)** and **(6, 0)** #### (b) *y-axis* (\( x = 0 \)): - \( -2y \leq 36 \implies y \geq -18 \) (but \( y \geq 0 \) by constraint) - \( 8y \leq 40 \implies y \leq 5 \) So, possible points: **(0, 0)** and **(0, 5)** --- ### Step 2: **Intersection of the Two Lines** Solve: \[ 6x - 2y = 36 \\ -x + 8y = 40 \] Multiply the second equation by 6 for elimination: \[ -6x + 48y = 240 \] Now add to the first equation: \[ 6x - 2y + (-6x + 48y) = 36 + 240 \\ 46y = 276 \implies y = 6 \] Substitute \( y = 6 \) into the first equation: \[ 6x - 2(6) = 36 \implies 6x - 12 = 36 \implies 6x = 48 \implies x = 8 \] So, intersection at **(8, 6)** --- ## 3. **Test Which Points Are Feasible** Check all points found so far: - (0, 0) - (6, 0) - (0, 5) - (8, 6) Test each point in all constraints: | Point | \(6x-2y\leq36\) | \(-x+8y\leq40\) | \(x\geq0\) | \(y\geq0\) | Feasible? | |---------|----------------|----------------|------------|------------|------------| | (0,0) | 0 ≤ 36 | 0 ≤ 40 | Yes | Yes | Yes | | (6,0) | 36 ≤ 36 | -6 ≤ 40 | Yes | Yes | Yes | | (0,5) | -10 ≤ 36 | 40 ≤ 40 | Yes | Yes | Yes | | (8,6) | 48-12=36 ≤ 36 | -8+48=40 ≤ 40 | Yes | Yes | Yes | All four points are feasible. --- ## 4. **List the Corner Points** **Corner points of the feasible region:** \[ \boxed{(0,0), (6,0), (0,5), (8,6)} \] --- ## 5. **Objective Function:** \[ z = -6x + 4y \] --- ## 6. **Test Corner Points in the Objective Function** Calculate \(z\) at each corner: - At (0,0): \(z = -6(0) + 4(0) = 0\) - At (6,0): \(z = -6(6) + 4(0) = -36\) - At (0,5): \(z = -6(0) + 4(5) = 20\) - At (8,6): \(z = -6(8) + 4(6) = -48 + 24 = -24\) --- ## 7. **Find the Maximum Value** - (0,0): 0 - (6,0): -36 - (0,5): 20 **(Maximum)** - (8,6): -24 So, the **maximum value is 20 at point (0, 5)**. --- # **Final Answers** ### **Corner points of the feasible region:** \[ \boxed{(0,0), (6,0), (0,5), (8,6)} \] ### **The "winning" corner point is:** \[ \boxed{(0,5)} \] ### **The maximum value of the objective function is:** \[ \boxed{20} \] --- Let me know if you need the graph or further breakdown!