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Solve the following linear programming problem graphically. Maximise Z = 4x1 + x2 subject to the constraints x1 + x2 ≤ 50; 3x1 + x2 ≤ 90 and x1 ≥ 0, x2 ≥ 0. - Business Mathematics and Statistics

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प्रश्न

Solve the following linear programming problem graphically.

Maximise Z = 4x1 + x2 subject to the constraints x1 + x2 ≤ 50; 3x1 + x2 ≤ 90 and x1 ≥ 0, x2 ≥ 0.

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उत्तर

Maximise Z = 4x1 + x2

Subject to the constraints

x1 + x2 ≤ 50

3x1 + x2 ≤ 90

x1, x2 ≥ 0

Since the decision variables, x1 and x2 are non-negative, the solution lies in the I quadrant of the plane.

Consider the equations

x1 + x2 = 50

x1 0 50
x2 50 0

3x1 + x2 = 90

x1 0 30
x2 90 0

The feasible region is OABC and its co-ordinates are O(0, 0) A(30, 0) C(0, 50) and B is the point of intersection of the lines

x1 + x2 = 50 ..........(1)

and 3x1 + x2 = 90 .........(2)

Verification of B:

x1 + x2 = 50 ..........(1)
3x1 + x2 = 90 .........(2)
−     −       −       
− 2x1 = − 40

x1 = 20

From (1), 20 + x2 = 50

x2 = 30

∴ B is (20, 30)

Corner points Z = 4x1 + x2
O(0, 0) 0
A(30, 0) 120
B(20, 30) 110
C(0, 50) 50

Maximum value occurs at A(30, 0)

Hence the solution is x1 = 30, x2 = 0 and Zmax = 120.

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अध्याय 10: Operations Research - Miscellaneous Problems [पृष्ठ २५२]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 11 TN Board
अध्याय 10 Operations Research
Miscellaneous Problems | Q 3 | पृष्ठ २५२

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