Algebra

Linear Programming

Easy

core

The feasible region represented by the constraints $4 x+y \geq 80, x+5 y \geq 115,3 x+2 y \leq 150, x, y \geq 0$ of an LPP is

Region A

Region B

Region C

Region D

✅ Correct Option: 3

The constraints for the linear programming problem are:

$4x + y \geq 80$

$x + 5y \geq 115$

$3x + 2y \leq 150$

$x, y \geq 0$

To identify the region, find the intercepts for each boundary line:

For $4x + y = 80$ :

When $x = 0, y = 80$

When $y = 0, x = 20$

For $x + 5y = 115$ :

When $x = 0, y = 23$

When $y = 0, x = 115$

For $3x + 2y = 150$ :

When $x = 0, y = 75$

When $y = 0, x = 50$

Test the origin $(0,0)$ against the inequalities to determine the direction of the feasible area:

$4(0) + (0) \geq 80 \implies 0 \geq 80$ (False; region is away from the origin)

$0 + 5(0) \geq 115 \implies 0 \geq 115$ (False; region is away from the origin)

$3(0) + 2(0) \leq 150 \implies 0 \leq 150$ (True; region is toward the origin)

The intersection of these half-planes, constrained by $x, y \geq 0$ in the first quadrant, forms the bounded polygonal area labeled as Region C.

2025: 16 May Shift 1

The corner points of the bounded feasible region determined by a set of constraints (linear inequalities) are A(0, 5), B(3, 5), C(5, 0) and D(4, 1) and the objective function is z = $px$ + 2 $qy$ where p,q > 0. The condition on $p$ and $q$ such that the maximum z occurs at B and D, is:

2025: 14 May Shift 2

Consider the linear programming problem(LPP):

Minimize $Z = x + y$

$x + 2y \leq 4,$

$3x + y \geq 3,$

$4x + 3y \geq 6,$

$x, y \geq 0.$

Which of the following is correct for the above linear programming problem (LPP):

(A) The LPP has a bounded feasible region.

(B) The LPP has a unique optimal solution.

(D) The corner points of the feasible region are (3/2, 0), (3/5, 6/5), (2/5, 6/5) and (4, 0)

Choose the correct answer from the options given below:

2026: 18 May Shift 1

The corner points of the bounded feasible region determined by $2x+y \leq 10$ , $x+3y \leq 15$ , $x,y \geq 0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let $z = px+qy$ , $p,q > 0$ , then the relation between $p$ and $q$ so that the maximum of $z$ occurs at both (3, 4) and (0, 5), is

2025: 22 May Shift 2

If the minimum value of the objective function $Z = ax + by$ of an LPP occurs at two points $(3, 5)$ and $(5, 3)$ , then

2025: 13 May Shift 2

The objective function of an LPP is $z = ax + by$ . If the maximum value of the objective function is 180, which occurs at two points (15,15) and (0,20), then which one of the following is true?

2025: 14 May Shift 1

Consider the LPP: Minimize $Z = x + 2y$ subject to $2x + y \geq 3$ , $x + 2y \geq 6$ , $x, y \geq 0$ . The optimal feasible solution occurs at

The feasible region represented by the constraints 4x+y≥80,x+5y≥115,3x+2y≤150,x,y≥04 x+y \geq 80, x+5 y \geq 115,3 x+2 y \leq 150, x, y \geq 04x+y≥80,x+5y≥115,3x+2y≤150,x,y≥0 of an LPP is

Solution

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The feasible region represented by the constraints $4 x+y \geq 80, x+5 y \geq 115,3 x+2 y \leq 150, x, y \geq 0$ of an LPP is