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CBSE Class 12 · 2017 · Linear ProgrammingLong Answer

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

Z = 5 x + 10 y

subject to the constraints

x + 2 y \leq 120

x + y \geq 60

x - 2 y \geq 0

x \geq 0, y \geq 0

Answer: Min Z = 300 at (60, 0)

Solution

Boundary lines:

x + 2 y = 120

,

x + y = 60

and

x - 2 y = 0

.
Corner points of the feasible region:

A (60, 0)

,

B (120, 0)

,

C (60, 30)

(intersection of

x + 2 y = 120

and

x - 2 y = 0

) and

D (40, 20)

(intersection of

x + y = 60

and

x - 2 y = 0

).
Evaluate

Z = 5 x + 10 y

:

A (60, 0)

:

Z = 300

← minimum

B (120, 0)

:

Z = 600

C (60, 30)

:

Z = 300 + 300 = 600

D (40, 20)

:

Z = 200 + 200 = 400

∴ Z_{m i n} = 300 at (60, 0)

Structure Feasible region for x + 2y = 120, x + y = 60 and x - 2y = 0, with corner points A(60, 0), B(120, 0), C(60, 30) and D(40, 20). — Feasible region for x + 2y = 120, x + y = 60 and x - 2y = 0, with corner points A(60, 0), B(120, 0), C(60, 30) and D(40, 20).

CBSE Class 12 Board · 2017 · Subjective

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