Question

Solve the following Linear Programming Problem graphically :
Maximise \(Z = 200x + 120y\)
subject to the constraints
\(x + y \le 300\)
\(3x + y \le 600\)
\(x - y \ge -100\)
\(x, y \ge 0\)

Hint:

To find intersection points like (150, 150), solve the pair of linear equations simultaneously.
The constraint \(x - y \ge -100\) can be written as \(y - x \le 100\). Always test \((0,0)\) to confirm the shading direction.

Answer 1

Concept:
• Graph the boundary lines for each constraint.
• Shade the feasible region (common region satisfying all inequalities).
• Corner Point Method: Maximum value must occur at one of the vertices of the feasible region.

Step 1: Determine intercepts and plot constraint lines
\(L_1: x + y = 300\). Intercepts: \((300, 0)\) and \((0, 300)\).
\(L_2: 3x + y = 600\). Intercepts: \((200, 0)\) and \((0, 600)\).
\(L_3: x - y = -100\). Intercepts: \((-100, 0)\) (outside 1st quad) and \((0, 100)\).
For \(L_3\), another point in the 1st quadrant: when \(x = 100, y = 200\).

Step 2: Identify the feasible region and corner points
\includegraphics[width=1\linewidth]{Q29_sol.png} The feasible region is bounded by corner points found by intersecting lines:
\(O: (0, 0)\)
\(A: (200, 0)\) (from \(L_2\) and x-axis)
\(B: (150, 150)\) (intersecting \(L_1\) and \(L_2\))
\(C: (100, 200)\) (intersecting \(L_1\) and \(L_3\))
\(D: (0, 100)\) (from \(L_3\) and y-axis)

Step 3: Evaluate \(Z = 200x + 120y\) at each corner point
At \(O(0, 0): Z = 0\)
At \(A(200, 0): Z = 200(200) + 0 = 40,000\)
At \(B(150, 150): Z = 200(150) + 120(150) = 30,000 + 18,000 = 48,000\)
At \(C(100, 200): Z = 200(100) + 120(200) = 20,000 + 24,000 = 44,000\)
At \(D(0, 100): Z = 200(0) + 120(100) = 12,000\) The maximum value of \(Z\) is \(48,000\) at point \((150, 150)\).