Question

Solve the following linear programming problem graphically:

Minimise \( z = 5x - 2y \)

Subject to the constraints:

\[ x + 2y \leq 120, \\ x + y \geq 60, \\ x - 2y \geq 0, \\ x \geq 0, \\ y \geq 0. \]

Hint:

In graphical linear programming, identify the feasible region, locate the corner points, and evaluate the objective function at each corner point to find the optimal solution.

Answer 1

To solve this linear programming problem graphically, follow these steps:

Step 1: Plot the constraints.
\( x + 2y \leq 120 \): Rewrite as \( y \leq \frac{120 - x}{2} \). This is a line passing through \( (0, 60) \) and \( (120, 0) \).
\( x + y \geq 60 \): Rewrite as \( y \geq 60 - x \). This is a line passing through \( (0, 60) \) and \( (60, 0) \).
\( x - 2y \geq 0 \): Rewrite as \( y \leq \frac{x}{2} \). This is a line passing through \( (0, 0) \) and \( (120, 60) \).
\( x \geq 0, y \geq 0 \): These are the \( x \)- and \( y \)-axes, restricting the feasible region to the first quadrant.

Step 2: Identify the feasible region.

The feasible region is the area that satisfies all the constraints. Plot the lines and shade the overlapping region that satisfies the inequalities.

Step 3: Determine the corner points of the feasible region.

The corner points are the intersections of the constraint lines:

\[ \text{Solve } x + 2y = 120 \text{ and } x + y = 60. \]

\[ y = 60, \quad x = 0 \Rightarrow (0, 60). \]

\[ \text{Solve } x + y = 60 \text{ and } x - 2y = 0. \]

\[ 2y + y = 60 \Rightarrow y = 20, \quad x = 40 \Rightarrow (40, 20). \]

\[ \text{Solve } x - 2y = 0 \text{ and } x + 2y = 120. \]

\[ 2y + 2y = 120 \Rightarrow y = 30, \quad x = 60 \Rightarrow (60, 30). \]

Step 4: Evaluate the objective function at each corner point.

\[ \text{At } (0, 60): \quad z = 5(0) - 2(60) = -120. \]

\[ \text{At } (40, 20): \quad z = 5(40) - 2(20) = 200 - 40 = 160. \]

\[ \text{At } (60, 30): \quad z = 5(60) - 2(30) = 300 - 60 = 240. \]

Step 5: Conclusion.

The minimum value of \( z = 5x - 2y \) occurs at \( (0, 60) \) with \( z = -120 \).

Final Answer:

\[ \boxed{z_{\text{min}} = -120 \text{ at } (0, 60)} \]