Algebra

Linear Programming

Medium

common

The feasible region represented by the constraints $x + y \leq 50, 3x + y \leq 90, x \geq 0, y \geq 0$ of an LPP is

Region D

Region B

Region A

Region C

✅ Correct Option: 3

The constraints are:

$x + y \leq 50$

$3x + y \leq 90$

$x \geq 0,\ y \geq 0$

The feasible region is the area where all constraints are satisfied simultaneously. Since $x \geq 0$ and $y \geq 0$ , the region is restricted to the first quadrant.

For $x + y = 50$ , the line passes through $(50,\ 0)$ and $(0,\ 50)$ .

Since $x + y \leq 50$ , the feasible side is towards the origin.

For $3x + y = 90$ , the line passes through $(30,\ 0)$ and $(0,\ 90)$ .

Since $3x + y \leq 90$ , the feasible side is towards the origin.

The intersection of the two lines:

$x + y = 50 \quad ...(i)$

$3x + y = 90 \quad ...(ii)$

$(ii) - (i)$ :

$2x = 40$

$x = 20$

$y = 50 - 20 = 30$

Intersection point $= (20,\ 30)$

The feasible region is bounded by the vertices:

$(0,\ 0)$ — Origin

$(30,\ 0)$ — $3x + y = 90$ meets the $x$ -axis

$(20,\ 30)$ — Both lines intersect

$(0,\ 50)$ — $x + y = 50$ meets the $y$ -axis

This is the region closest to the origin in the first quadrant, where both shaded areas overlap, which corresponds to Region A.

$\boxed{\text{Option 3: Region A}}$

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The feasible region represented by the constraints x+y≤50,3x+y≤90,x≥0,y≥0x + y \leq 50, 3x + y \leq 90, x \geq 0, y \geq 0x+y≤50,3x+y≤90,x≥0,y≥0 of an LPP is

Solution

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The feasible region represented by the constraints $x + y \leq 50, 3x + y \leq 90, x \geq 0, y \geq 0$ of an LPP is