Applied Mathematics

Linear Programming

Medium

applied

If the objective function $z=10(x-7y+190)$ and the corner points of the bounded feasible region are A(0,4), B(0,5), C(3,5), D(5,3), E(5,0), F(4,0), then the minimum value of z will be:

1620

1550

1430

1580

✅ Correct Option: 2

Minimising $z$ means minimising $x-7y$ .

Values at the corners: $-28,\,-35,\,-32,\,-16,\,5,\,4$ , least at B $(0,5)$ .

Then $z=10(-35+190)=1550$ .

2023: 11 Aug Shift 1

Objective function of a LPP represent

2023: 7 June Shift 1

The Solution set of the inequality three $3x + 4y \leq 12$ is :

2022: 25 May Shift 1

The maximum value of $Z = 3x + y$ subject to the constraints $x + y \leq 30, 2x + y \leq 40, x, y \geq 0$ is

2022: 25 May Shift 1

Match List I with List II

LIST I	LIST II
A. A solution that does not satisfy all the constraints is called	I. Linear
B. The objective function in an LPP is	II. Convex polygon
C. Linear inequalities or equations on the variables of LPP are called	III. Infeasible solution
D. The feasible region in an LPP, formed by the convex combinations of the corner points, is called	IV. Constraints

Choose the correct answer from the options given below:

2023: 22 May Shift 3

If the objective function for an L.P.P. is $z = 3x + 4y$ and the corner points for unbounded feasible region are (9, 0), (4, 3), (2, 5) and (0, 8), then the minimum value of $z$ occurs at :

2023: 11 Aug Shift 1

For the LPP, Min $Z = 6x + 10y$ subject to $x \geq 6, y \geq 3, 2x + y \geq 10, x \geq 0, y \geq 0$ , redundant constraint is :

If the objective function z=10(x−7y+190)z=10(x-7y+190)z=10(x−7y+190) and the corner points of the bounded feasible region are A(0,4), B(0,5), C(3,5), D(5,3), E(5,0), F(4,0), then the minimum value of z will be:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If the objective function $z=10(x-7y+190)$ and the corner points of the bounded feasible region are A(0,4), B(0,5), C(3,5), D(5,3), E(5,0), F(4,0), then the minimum value of z will be: