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Linear Programming — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsLinear Programming4 Marks
Question
Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when relationship is expressed as linear equations or inequations.
Based on the above information, answer the following questions.
  1. The optimal value of the objective function is attained at the points:
  1. On X-axis.
  2. On Y-axis.
  3. Which are comer points of the feasible region.
  4. None of these.
  1. The graph of the inequality 3x + 4y < 12 is:
  1. Half plane that contains the origin.
  2. Half plane that neither contains the origin nor the points of the line 3x + 4y = 12.
  3. Whole XOY-plane excluding the points on line 3x + 4y = 12.
  4. None of these.
  1. The feasible region for an LPP is shown in the figure. Let Z = 2x + 5y be the objective function. Maximum of Z occurs at:
  1. (7, 0)
  2. (6, 3)
  3. (0, 6)
  4. (4, 5)
  1. The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points ( 15, 15) and (0, 20) is:
  1. p = q
  2. p = 2q
  3. q = 2p
  4. q = 3p
  1. The comer points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.
Compare the quantity in Column A and Column B
Column A
Column B
Maximum of Z
325
  1. The quantity in column A is greater.
  2. The quantity in column Bis greater.
  3. The two quantities are equal.
  4. The relationship cannot be determined on the basis of the information supplied.
✓

Answer

  1. (c) Which are comer points of the feasible region.
Solution:

When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at comer points of the feasible region.
  1. (d) None of these.
Solution:

From the graph of 3x + 4y < 12, it is clear that it contains the origin but not the points on the line 3x + 4y = 12.

  1. (d) (4, 5)
Solution:

Maximum of objective function occurs at corner points.
Corner Points
Value of Z = 2x + 5y
(0, 0)
0
(7, 0)
14
(6, 3)
27
(4, 5)
$33\leftarrow\text{Maximum}$
(0, 6)
30
  1. (d) q = 3p
​​​​​​​​​​​​​​Solution:

Value of Z = px + qy at ( 15, 15)= 15p + 15q and that at (0, 20) = 20q. According to given condition, we have 15p + 15q = 20q ⇒ 15p = Sq ⇒ q = 3p.
  1. (b) The quantity in column Bis greater.
​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​Solution:

Construct the following table of values of the objective function:
Corner Points
Value of Z = 4x + 3y
(0, 0)
4 × 0 + 3 × 0 = 0
(0, 40)
4 × 0 + 3 × 40 = 120
(20, 40)
4 × 20 + 3 × 40 = 200
(60, 20)
$4\times60+3\times20=300\leftarrow\text{Maximum}$
(60, 0)
4 × 60 + 3 × 0 = 240

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