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Linear Inequalities

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Question : 24 of 54

Marks: +1, -0

Graph of some constraints are given by the figure

L_{1}: x+2y=10 ; \;\; L_{2}=x+y=6 ; \;\; L_{3}: x-y=2 ; L_{4}: x-2y=1

The shaded region belongs to the constraints given by :

$x+2y \leq 10$
$x+y \geq 6$
$x-y \leq 2$
$x-2y \leq 1$
$x \geq 0 ; y \geq 0$
$x+2y \leq 10$
$x+y \leq 6$
$x-y \leq 2$
$x-2y \leq 1$
$x \geq 0 ; y \geq 0$
$x+2y \leq 10$
$x+y \leq 6$
$x-y \geq 2$
$x-2y \leq 1$
$x \geq 0 ; y \geq 0$
none of these

Solution:

Concept:

The feasible region (shaded area) is the intersection of half-planes defined by the constraints, determined by the side of each line that contains the region. From the figure, all lines bound the region from above (or below) as indicated by the shading direction.

Explanation:

Since the shading is on the side that includes the origin for all lines (typical bounded region in first quadrant), the inequalities for

L_1

,

L_2

,

L_3

, and

L_4

must all be "≤" (i.e., below or on each line).

Thus the constraints are:

x+2y \leq 10

,

x+y \leq 6

,

x-y \leq 2

,

x-2y \leq 1

, and

x \geq 0

,

y \geq 0

.

This matches Option B.

Answer:

Option B.

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