हिंदी

Show that the minimum of Z occurs at more than two points. Minimise and Maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

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प्रश्न

Show that the minimum of Z occurs at more than two points.

Minimise and Maximise Z = x + 2y 

subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

योग
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उत्तर

The system of constraints is:

x + 2y ≥ 100           ....(i)

2x - y ≤ 0                ....(ii)

2x + y ≤ 200           ....(iii)

and x, y ≥ 0           ....(iv)

Let l1 : x + 2y = 100

l2 : 2x - y = 0

l3 : 2x + y = 200

The shaded region in the figure is the feasible region determined by the system of constraints (i) to (iv).

It is observed that the feasible region ECDB is bounded.

Thus, we use the Corner Point Method to determine the maximum and minimum values of Z.

We have, Z = x + 2y

The co-ordinates of E, C, D and B are (20, 40) (on solving x + 2y = 100 and 2x - y = 0), 

(50, 100) (on solving 2x + y = 200 and 2x - y = 0), (0, 200) and (0, 50) respectively.

Corner Point Corresponding values of Z
(20, 40) 100
(50, 100) 250
(0, 200) 400 (Maximum)
(0, 50) 100

Hence Zmax = 400 at (0, 200) and Zmin = 100 at all points on the line segment joining the points (0, 50) and (20, 40).

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अध्याय 12: Linear Programming - Exercise 12.1 [पृष्ठ ५१४]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 12 Linear Programming
Exercise 12.1 | Q 8 | पृष्ठ ५१४

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