📐 Mathematics · Linear Programming · JEE

Maximize Z = 50x + 60y subject to 2x + y <= 120, x + 2y <= 120, x >= 0, y >= 0. The corner points of the feasible region are (0,0), (60,0), (40,40), and (0,60). What is the maximum value of Z?

Answer: 4400.

A 4400
B 3600
C 3000
D 6000

Correct answer: A. 4400

Explanation: Z(0,0)=0, Z(60,0)=3000, Z(40,40)=50(40)+60(40)=2000+2400=4400, Z(0,60)=3600. Maximum is 4400 at (40,40).

The feasible region (shaded) is bounded by the constraint lines; the fundamental theorem of LPP guarantees the optimal value of the objective function occurs at one of the corner points (O, A, B, C), so only these need to be checked, not the entire region.

Concept context

Optimizing a linear objective function subject to linear constraints using the graphical corner point method.

Read the full Linear Programming notes →

Maximize Z = 50x + 60y subject to 2x + y <= 120, x + 2y <= 120, x >= 0, y >= 0. The corner points of the feasible region are (0,0), (60,0), (40,40), and (0,60). What is the maximum value of Z?

Concept context

More Linear Programming questions