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Linear Programming — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsLinear Programming4 Marks
Question
Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region. Based on the above information, answer the following questions.
  1. Objective function of a L.P.P. is:
  1. A constant.
  2. A function to be optimised.
  3. A relation between the variables.
  4. None of these.
  1. Which of the following statement is correct?
  1. Every LPP has at least one optimal solution.
  2. Every LPP has a unique optimal solution.
  3. If an LPP has two optimal solutions, then it has infinitely many solutions.
  4. None of these.
  1. In solving the LPP: "minimize f = 6x + 10y subject to constraints $\text{x}\geq6,\text{ y}\geq2,\text{ 2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0"$ redundant constraints are:
  1. $\text{x}\geq6,\text{ y}\geq2$
  2. $\text{2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0$
  3. $\text{x}\geq6$
  4. None of these
  1. The feasible region for a LPP is shown shaded in the figure. Let Z = 3x - 4y be the objective function. Minimum of Z occurs at:
  1. (0, 0)
  2. (0, 8)
  3. (5, 0)
  4. (4, 10)
  1. The feasible region for a LPP is shown shaded in the figure. Let F = 3x - 4y be the objective function. Maximum value of F is:
  1. 0
  2. 8
  3. 12
  4. -18
✓

Answer

  1. (b) A function to be optimised.
Solution:

Objective function is a linear function (involve variable) whose maximum or minimum value is to be found.
  1. (c) If an LPP has two optimal solutions, then it has infinitely many solutions.
Solution:

If optimal solution is obtained at two distinct points A and B ( corners of the feasible region), then optimal solution is obtained at every point of segment [AB].
  1. (b) $\text{2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0$
​​​​​​​Solution:

When $\text{x}\geq6$ and $\text{y}\geq2,$ then

$\text{2x}+\text{y}\geq2\times6+2,\text{i.e.,}\text{ 2x}+\text{y}\geq14$

Hence, $\text{x}\geq0,\text{ y}\geq0$ and $2\text{x}+\text{y}\geq10$ are automatically satisfied by every point of the region

$\{(\text{x, y}):\text{x}\geq6\}\cap\{(\text{x, y}):\text{y}\geq2\}$
  1. (b) (0, 8)
​​​​​​​​​​​​​​​​​​​​​Solution:

Construct the following table of values of the objective function:
Corner Point
Value of Z = 3x - 4y
(0, 0)
3 × 0 - 4 × 0 = 0
(5, 0)
3 × 5 - 4 × 0 = 15
(6, 5)
3 × 6 - 4 × 5 = -2
(6, 8)
3 × 6 - 4 × 8 = -14
(4, 10)
3 × 4 - 4 × 10 = -28
(0, 8)
$3\times0 - 4\times8 = -32\leftarrow\text{Minimum}$
Minimum of Z = -32 at (0, 8).
  1. (a) 0
​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Construct the following table of values of the objective function F:
Corner Point
Value of F = 3x - 4y
(0, 0)
$3\times0 - 4\times0 = 0\leftarrow\text{Minimum}$
( 6, 12)
3 × 6 - 4 × 12 = -30
(6, 16)
3 × 6 - 4 × 16 = -46
(0, 4)
3 × 0 - 4 × 4 = -16
Hence, maximum of F = 0.

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