CBSE Class 12 · 2005 · Linear ProgrammingLong Answer

Solve the following linear programming problem graphically:
Minimize

Z = 3 x + 5 y

subject to the constraints

x + y \geq 2

x + 3 y \geq 3

x \geq 0, y \geq 0

Answer: Min Z = 7 at (3/2, 1/2)

Solution

Boundary lines:

x + y = 2

and

x + 3 y = 3

.
The feasible region is unbounded. Its corner points are

(3, 0)

(

x + 3 y = 3

on the

x

-axis),

(\frac{3}{2}, \frac{1}{2})

(intersection) and

(0, 2)

(

x + y = 2

on the

y

-axis).
Intersection: subtract

x + y = 2

from

x + 3 y = 3

to get

2 y = 1 ⟹ y = \frac{1}{2}

, so

x = \frac{3}{2}

.
Evaluate

Z = 3 x + 5 y

(3, 0)

Z = 9

(\frac{3}{2}, \frac{1}{2})

Z = \frac{9}{2} + \frac{5}{2} = 7

← minimum

(0, 2)

Z = 10

Since the region is unbounded, check the open half-plane

3 x + 5 y < 7

: it has no point in common with the feasible region, so the minimum is genuine.

∴ Z_{m i n} = 7 at (\frac{3}{2}, \frac{1}{2})

CBSE Class 12 Board · 2005 · Subjective