Q. 44.0( 3 Votes )

# Solve the following L.P.P. graphically :

Minimize

Subject to

Constraints

and *[CBSE 2017]*

*[CBSE 2017]*

Answer :

Minimize

Z = 5x + 10y

Subject to constraints

x + 2y ≤ 120

x + y ≥ 60

x – 2y ≥ 0

and x , y ≥ 0

Firstly, we draw the graph of the line x + 2y = 120

Put (0,0) in the inequality x + 2y ≤ 120, we get

0 + 2 × 0 ≤ 120

⇒ 0 ≤ 120 (which is true)

So, half plane is towards the origin.

Secondly, graph of the line x + y = 60

Put (0,0) in the inequality x + y ≥ 60, we get

0 + 0 ≥ 60

⇒ 0 ≥ 60 (which is false)

So, half plane is away from the origin.

Thirdly, draw the graph of line x – 2y = 0

On solving equations x – 2y = 0 and x + y = 60, we get B(40, 20)

and on solving equations x – 2y = 0 and x + 2y = 120, we get C(60, 30).

Feasible Region is ABCDA.

So, the minimum value of Z is 300 at the point (60, 0)

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Using the method of integration, find the area of the region bounded by the following lines:

5x - 2y - 10 = 0

x + y - 9 = 0

2x - 5y - 4 = 0

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