CBSE Class 12 · 2023 · Linear ProgrammingLong Answer

Solve the following linear programming problem graphically:
Maximize

Z = x + 2 y

subject to the constraints

x + 2 y \geq 100

2 x - y \leq 0

2 x + y \leq 200

x \geq 0, y \geq 0

Answer: Max Z = 400 at (0, 200)

Solution

Convert the inequalities to boundary lines:

x + 2 y = 100

2 x - y = 0

and

2 x + y = 200

.
The feasible corner points are:

A (0, 50)

—

x + 2 y = 100

meets

x = 0

B (20, 40)

—

x + 2 y = 100

meets

2 x - y = 0

C (50, 100)

—

2 x - y = 0

meets

2 x + y = 200

D (0, 200)

—

2 x + y = 200

meets

x = 0

Evaluate

Z = x + 2 y

A (0, 50)

Z = 100

B (20, 40)

Z = 100

C (50, 100)

Z = 250

D (0, 200)

Z = 400

← maximum

∴ Z_{m a x} = 400 at (0, 200)

Structure Feasible region bounded by x + 2y = 100, 2x - y = 0 and 2x + y = 200, with corner points A(0, 50), B(20, 40), C(50, 100) and D(0, 200). — Feasible region bounded by x + 2y = 100, 2x - y = 0 and 2x + y = 200, with corner points A(0, 50), B(20, 40), C(50, 100) and D(0, 200).

CBSE Class 12 Board · 2023 · Subjective

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