Question

Find the minimum value of \[ Z = 50x + 70y \] \(\text{under the following constraints by graphical method:}\) \[ 2x + y \geq 8, \] \[ x + 2y \geq 10, x \geq 0, y \geq 0. \]

Hint:

In the graphical method for linear programming, plot the constraints, identify the feasible region, and evaluate the objective function at the vertices of the feasible region to find the optimal solution.

Answer 1

To solve this linear programming problem by the graphical method, we follow these steps: 

Step 1: Graph the Constraints We first graph the constraints as equations and find the feasible region. 

1. For the constraint \( 2x + y \geq 8 \), we convert it to the equality form: \[ 2x + y = 8. \] 

To graph this line, find two points: 

- When \( x = 0 \), \( y = 8 \), so one point is \( (0, 8) \). 

- When \( y = 0 \), \( 2x = 8 \), so \( x = 4 \), and the other point is \( (4, 0) \). 

2. For the constraint \( x + 2y \geq 10 \), we convert it to the equality form: \[ x + 2y = 10. \] 

To graph this line, find two points: 

- When \( x = 0 \), \( 2y = 10 \), so \( y = 5 \), and one point is \( (0, 5) \). 

- When \( y = 0 \), \( x = 10 \), and the other point is \( (10, 0) \). 

3. For the constraints \( x \geq 0 \) and \( y \geq 0 \), the feasible region is limited to the first quadrant. 

Step 2: Plot the Lines Plot the lines on the graph and identify the feasible region that satisfies all constraints. This will be a polygon with vertices at the points of intersection of the lines. 

Step 3: Find the Intersection Points We need to find the intersection points of the lines: 1. Solve \( 2x + y = 8 \) and \( x + 2y = 10 \) simultaneously: \[ 2x + y = 8 \text{(i)} \] \[ x + 2y = 10 \text{(ii)}. \] From (ii), \( x = 10 - 2y \). Substitute this into (i): \[ 2(10 - 2y) + y = 8, \] \[ 20 - 4y + y = 8, \] \[ -3y = -12, \] \[ y = 4. \] Substitute \( y = 4 \) into \( x + 2y = 10 \): \[ x + 2(4) = 10 \Rightarrow x + 8 = 10 \Rightarrow x = 2. \] Therefore, the intersection point is \( (2, 4) \). 

Step 4: Evaluate the Objective Function at the Vertices The vertices of the feasible region are the points of intersection and the intercepts with the axes. We evaluate \( Z = 50x + 70y \) at each of these points. 

- At \( (0, 8) \): \[ Z = 50(0) + 70(8) = 560. \] - At \( (4, 0) \): \[ Z = 50(4) + 70(0) = 200. \] - At \( (10, 0) \): \[ Z = 50(10) + 70(0) = 500. \] - At \( (2, 4) \): \[ Z = 50(2) + 70(4) = 100 + 280 = 380. \] Step 5: Find the Minimum Value From the above evaluations, the minimum value of \( Z \) occurs at \( (4, 0) \), where \( Z = 200 \). 

Conclusion: The minimum value of \( Z = 50x + 70y \) is \[ \boxed{200}, \] which occurs at \( (x, y) = (4, 0) \).