The maximum value of \( Z = 4x + y \) for a L.P.P. whose feasible region is given below is:
Show Hint
- \( 50 \)
- \( 110 \)
- \( 120 \)
- \( 170 \)
The Correct Option is C
Solution and Explanation
Step 1: {Identify the corner points of the feasible region}
From the graph, the vertices of the feasible region are: \[ A(0, 50), \, B(20, 30), \, C(30, 0). \]
Step 2: {Substitute corner points into \( Z = 4x + y \)}
Evaluate \( Z \) at each vertex: \[ Z_A = 4(0) + 50 = 50,\] \[\quad Z_B = 4(20) + 30 = 110,\]
\(\quad Z_C = 4(30) + 0 = 120.\)
Step 3: {Find the maximum value}
The maximum value of \( Z \) occurs at \( C(30, 0) \), where \( Z = 120 \).
Step 4: {Verify the options}
The maximum value is \( 120 \), which corresponds to option (C).
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