The corner points of the feasible region of the LPP: Minimize Z = -50x + 20y subject to 2x - y ≥ -5, 3x + y ≥ 3, 2x - 3y ≤ 12 and x, y ≥ 0 are Option 1: (0,5), (0,6), (1,0), (6,0) Option 2: (0,3), (0,5), (3,0), (6,0) Option 3: (0,3), (0,5), (1,0), (6,0) Option 4: (0,5), (0,6), (1,0), (3,0)

Answer: (0,3), (0,5), (1,0), (6,0). Solution: The corner points (vertices) of the feasible region are found where two constraint lines intersect or where a constraint line meets an axis. Given constraints: 2x - y ≤ -5 which gives y ≥ 2x + 5 3x + y ≥ 3 which gives y ≥ 3 - 3x 2x - 3y ≤ 12 which gives y ≥ (2x - 12)/3 x ≥ 0, y ≥ 0 For points on the y-axis, set x = 0: From y ≥ 2(0) + 5: y ≥ 5 Boundary point: (0, 5) From y ≥ 3 - 3(0): y ≥ 3 Boundary point: (0, 3) From y ≥ (2(0) - 12)/3: y ≥ -4 This does not restrict y further in the first quadrant. For points on the x-axis, set y = 0: From 2x - 0 ≤ -5: x ≤ -2.5 This is not possible since x ≥ 0. From 3x + 0 ≥ 3: x ≥ 1 Boundary point: (1, 0) From 2x - 3(0) ≤ 12: x ≤ 6 Boundary point: (6, 0) The corner points of the feasible region are: (0, 3) (0, 5) (1, 0) (6, 0)

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Linear Programming

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The corner points of the feasible region of the LPP: Minimize $Z = -50x + 20y$ subject to $2x - y \geq -5$ , $3x + y \geq 3$ , $2x - 3y \leq 12$ and $x, y \geq 0$ are

(0,5), (0,6), (1,0), (6,0)

(0,3), (0,5), (3,0), (6,0)

(0,3), (0,5), (1,0), (6,0)

(0,5), (0,6), (1,0), (3,0)

✅ Correct Option: 3

The corner points (vertices) of the feasible region are found where two constraint lines intersect or where a constraint line meets an axis.

Given constraints:

$2x - y \leq -5$ which gives $y \geq 2x + 5$

$3x + y \geq 3$ which gives $y \geq 3 - 3x$

$2x - 3y \leq 12$ which gives $y \geq \frac{2x - 12}{3}$

$x \geq 0, y \geq 0$

For points on the y-axis, set $x = 0$ :

From $y \geq 2(0) + 5$ :

$y \geq 5$

Boundary point: $(0, 5)$

From $y \geq 3 - 3(0)$ :

$y \geq 3$

Boundary point: $(0, 3)$

From $y \geq \frac{2(0) - 12}{3}$ :

$y \geq -4$

This does not restrict $y$ further in the first quadrant.

For points on the x-axis, set $y = 0$ :

From $2x - 0 \leq -5$ :

$x \leq -2.5$

This is not possible since $x \geq 0$ .

From $3x + 0 \geq 3$ :

$x \geq 1$

Boundary point: $(1, 0)$

From $2x - 3(0) \leq 12$ :

$x \leq 6$

Boundary point: $(6, 0)$

The corner points of the feasible region are:

$(0, 3)$

$(0, 5)$

$(1, 0)$

$(6, 0)$

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Minimize $z = 2x - 5y$ , subject to the constraints:

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In reference to a linear programming problem (LPP), which of the following statements are true?

A. The optimal value of the objective function is attained at the points given by intersections of inequation with the axes only.

B. The objective function of a LPP is a linear function to be optimized.

C. Every LPP admits an optimal solution.

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Choose the correct answer from the options given below:

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The corner points of the feasible region of the LPP: Minimize Z=−50x+20yZ = -50x + 20yZ=−50x+20y subject to 2x−y≥−52x - y \geq -52x−y≥−5, 3x+y≥33x + y \geq 33x+y≥3, 2x−3y≤122x - 3y \leq 122x−3y≤12 and x,y≥0x, y \geq 0x,y≥0 are

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The corner points of the feasible region of the LPP: Minimize $Z = -50x + 20y$ subject to $2x - y \geq -5$ , $3x + y \geq 3$ , $2x - 3y \leq 12$ and $x, y \geq 0$ are