Linear Programming - Class 12 Mathematics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Linear Programming | NCERT | Mathematics | Class 12
Which of the following are feasible solutions for a linear programming problem with constraints?
$$ \begin{array}{l} x \geq 0, y \geq 0 \ 3x + 5y \leq 15 \ 5x + 2y \leq 10 \end{array} $$
A) $(1,-1)$
B) $(-3,5)$
C) $(0,0)$
D) $(2,0)$
The feasible solutions for this linear programming problem, considering the constraints, are:
- C) $(0,0)$
- D) $(2,0)$
These solutions are termed "feasible" because they meet the criteria established by the constraints of the problem. Both $x$ and $y$ must be non-negative, and they must satisfy the following inequalities: $$ 3x + 5y \leq 15 \ 5x + 2y \leq 10 $$
Evaluation of each option:
- A) $(1, -1)$: This does not satisfy the constraint $y \geq 0$.
- B) $(-3, 5)$: This violates the constraint $x \geq 0$.
-
C) $(0, 0)$: This point satisfies all constraints:
- $x = 0 \geq 0$ and $y = 0 \geq 0$
- $3(0) + 5(0) = 0 \leq 15$
- $5(0) + 2(0) = 0 \leq 10$
-
D) $(2, 0)$: This point also satisfies all constraints:
- $x = 2 \geq 0$ and $y = 0 \geq 0$
- $3(2) + 5(0) = 6 \leq 15$
- $5(2) + 2(0) = 10 \leq 10$
Conclusion: Options C and D are the correct feasible solutions because they adhere to all the constraints of the linear programming problem.
The corner points of a feasible region determined by a system of linear inequalities are $(20, 40)$, $(50, 100)$, $(0, 200)$, and $(0, 50)$. If the objective function $Z = x + 2y$, then the maximum of $Z$ occurs at:
A. $(20, 40)$
B. $-50100$
C. $(0, 200)$
D. $(0, 50)$
To determine where the objective function $Z = x + 2y$ reaches its maximum, the strategy is to evaluate $Z$ at all given corner points of the feasible region. These points are $(20, 40)$, $(50, 100)$, $(0, 200)$, and $(0, 50)$.
By calculating $Z$ using each point:
At point $(20, 40)$:$$ Z = 20 + 2 \times 40 = 20 + 80 = 100 $$
At point $(50, 100)$:$$ Z = 50 + 2 \times 100 = 50 + 200 = 250 $$
At point $(0, 200)$:$$ Z = 0 + 2 \times 200 = 0 + 400 = 400 $$
At point $(0, 50)$:$$ Z = 0 + 2 \times 50 = 0 + 100 = 100 $$
The maximum value of $Z = 400$ occurs at the point $(0, 200)$. Hence, the correct answer to the problem, where the objective function attains its maximum, is given by:
C. $(0, 200)$.
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