Linear Programming

The line L₁: y = x = 0 and L₂: 2x + y = 0 intersect the line L₃: y + 2 = 0 at P and Q respectively. The bisectorof the acute angle between L₁ and L₂ intersects L₃ at R.
Statement-1: The ratio PR: RQ equals 2√2:√5
Statement-2: In any triangle, the bisector of an angle divides the triangle into two similar triangles.

• Statement-1 is true, Statement-2 is true ; Statement-2 is correct explanation for Statement-1

• Statement-1 is true, Statement-2 is true ; Statement-2 is not a correct explanation for Statement-1

• Statement-1 is true, Statement-2 is false

• Statement-1 is true, Statement-2 is false

If x is real, the maximum value of 

• 1/4

• 41

• 1

• 1

The function  has a local minimum at

• x = 2

• x = −2

• x = 0

• x = 0

For the LPP Min z = x₁ + x₂ such that inequalities 5x₁ + 10x₂ $\ge$ 0, x₁ + x₂ $\le$ 1, x₂ $\le$ 4 and x₁, x₂ > 0

• There is a bounded solution

• There is no solution

• There are infinite solutions

• None of these

The maximum value of z = 9x + 13y subject to 2x + 3y $\le$ 18, 2x + y $\le$ 10, x $\ge$ 0, y $\ge$ 0 is

• 130

• 81

• 79

• 99

The maximum value of the objective function Z = 3x + 2y for linear constraints x + y $\le$ 7, 2x + 3y $\le$ 16, x² $\ge$ 0, y² $\ge$ 0 is

• 16

• 21

• 25

• 28

A diet of a sick person must contain atleast 4000 unit of vitamins, 50 unit of proteins and 1400 calories. Two foods A and B are available at cost rs. 4 and rs. 3 per unit respectively. If one unit of A contains 200 unit of vitamins, 1 unit of protein and 40 calories, while one unit of food B contains 100 unit of vitamins, 2 unit of protein and 40 calories. Formulate the problem, so that the diet be cheapest

• $$\begin{gathered} 200\mathrm{x} + 100\mathrm{y} \ge 4000, \mathrm{x} + 2\mathrm{y} \ge 50 \\ 40\mathrm{x} + 40\mathrm{y} \ge 1400, \mathrm{x} \ge 0 \mathrm{and} \mathrm{y} \ge 0 \\ \mathrm{O}. \mathrm{F} \mathrm{z} = 4\mathrm{x} + 3\mathrm{y} \end{gathered}$$

• $$\begin{gathered} 400\mathrm{x} + 200\mathrm{y} \ge 100, \mathrm{x} + 2\mathrm{y} \ge 50 \\ 40\mathrm{x} + 40\mathrm{y} \ge 1400, \mathrm{x} \ge 0 \mathrm{and} \mathrm{y} \ge 0 \\ \mathrm{O}. \mathrm{F} \mathrm{z} = 4\mathrm{x} + 3\mathrm{y} \end{gathered}$$

• $$\begin{gathered} 100\mathrm{x} + 200\mathrm{y} \ge 4000, \mathrm{x} + 2\mathrm{y} \ge 50 \\ 40\mathrm{x} + 40\mathrm{y} \ge 1400, \mathrm{x} \ge 0 \mathrm{and} \mathrm{y} \ge 0 \\ \mathrm{O}. \mathrm{F} \mathrm{z} = 4\mathrm{x} + 3\mathrm{y} \end{gathered}$$

• None of the above

The constraints - x₁ + x₂ $\le$ 1, - x₁ + 3x₂ $\le$ 9, x₁, x₂ > 0 defines on

• bounded feasible space

• unbounded feasible space

• both bounded and unbounded feasible space

• None of the above

The objective function Z = x₁ + x₂, subject to the constraints are  x₁ + x₂ $\le$ 10, - 2x₁ + 3x₂ $\le$ 15, x₁ $\le$ 6, x₁x₂ $\ge$ 0 has maximum value of _ the feasible region.

• at only one point

• at only two points

• at every point of the segment joining two points

• at every point of the line joining two points equivalent to

The objective function z = 4x₁ + 5x₂, subject to 2x₁ + x₂ $\ge$ 7, 2x₁ + 3x₂ $\le$ 15, x₂ $\le$ 3, x₁x₂ $\ge$ 0 has minimum value at the point

• on X-axis

• on Y-axis

• at the origin

• on the line parallel to X-axis