Class 12 Mathematics Chapter 12 Linear Programming

Key Concept:  Non-negative Constraints

a) $$x \geq 0, y \geq 0$$
[Solution Description] By definition, non-negative constraints ensure that variables cannot be negative, given as $$x \geq 0, y \geq 0$$.

Key Concept: Decision Variables

a) $$ x \text{ and } y $$
[Solution Description]
Decision variables are the unknowns that we are trying to solve for in a linear programming model. They represent choices available to the decision-maker.

Key Concept: Constraint Satisfaction

d) $$(10, 40)$$
[Solution Description]
Let’s check point $$(15, 30)$$ for satisfaction:

Investment constraint:

$$
2500(15) + 500(30) = 37500 + 15000 = 52500
$$

Since $$52500 > 50000$$, this does not satisfy.

Storage constraint:

$$
15 + 30 = 45 \leq 60
$$

Check point $$(20, 25)$$:

Investment constraint:

$$
2500(20) + 500(25) = 50000 + 12500 = 62500
$$

Since $$62500 > 50000$$, this does not satisfy.

Storage constraint:

$$
20 + 25 = 45 \leq 60
$$

Check point $$(10, 40)$$:

Investment constraint:

$$
2500(10) + 500(40) = 25000 + 20000 = 45000
$$

Since $$45000 \leq 50000$$, this satisfies.

Storage constraint:

$$
10 + 40 = 50 \leq 60
$$

Thus, the point $$(10, 40)$$ satisfies both constraints.

Key Concept: Decision Variables

a) x and y
[Solution Description] In this expression, $$x$$ and $$y$$ are the decision variables whose values determine the value of the objective function $$Z$$.

Key Concept:Complex Constraints, Real-world Application

a) Maximize $$Z = 30x + 20y$$

[Solution Description]

To formulate the problem:

Let x be the number of units of P1 produced.

Let y be the number of units of P2 produced.

Objective function: Maximize $$Z = 30x + 20y$$

Constraints:

$$2x + y \leq 40 \quad \text{(Time Constraint)}$$

$$x \leq 12 \quad \text{(Demand for P1)}$$

$$y \geq 10 \quad \text{(Demand for P2)}$$

$$x \geq 0,\; y \geq 0 \quad \text{(Non-negativity)}$$

Thus, the objective function considering all constraints is Maximize $$Z = 30x + 20y$$

Key Concept:  Basic Objective Function

a) Maximization
[Solution Description] The given objective function is $$Z = 4x + 6y$$. This type of linear function is often used to either maximize or minimize a certain output based on variables x and y. Here, since no specific direction is mentioned in the problem statement, we generally refer to maximization unless specified otherwise in many practical applications involving coefficients with positive values.

Key Concept:Constraints Analysis, Real-world Application

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] Both assertion and reason discuss aspects related to constraints and how they affect the feasible region in a linear programming problem. The storage constraints contribute to the boundaries forming the feasible region. However, the Reason mentions various constraints such as production limits and resource availability, which collectively form the feasible region through their intersection but do not solely relate to storage constraints.

The feasible region is indeed shaped by these intersections, yet the statement in the Reason does not correctly explain why storage constraints specifically would define this region’s particular shape distinctively.

Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Objective Function Basics

a) $$ Z = 7x + 4y $$
[Solution Description] The objective function in an LPP problem is the linear function that we aim to maximize or minimize. Here, it is given directly as $$Z = 7x + 4y$$.

Key Concept: Constraints Identification

b) Storage capacity cannot be exceeded
[Solution Description]
Constraints in a diet problem could include limits on resources such as budget, calories, or storage space. If storage space is limited, it becomes a constraint like $$\sum_{i} {s_ix_i} \leq S$$, where $$s_i$$ is the storage space for item $$i$$, $$x_i$$ is the amount of item $$i$$, and $$S$$ is the maximum available storage.

Key Concept:Objective Function Value

c) 6250

[Solution Description] Substitute the values of $$x = 10$$ and $$y = 50$$ into the objective function $$Z = 250x + 75y$$:

$$Z = 250(10) + 75(50) = 2500 + 3750 = 6250$$

The calculated value of the objective function Z at point $$(10, 50)$$ is $$6250$$.

Key Concept:Unbounded Region Analysis, Infeasible Solution Identification

d) If $$5x – 2y < M$$ exists $$\forall x,y$$ [Solution Description] For the given set of constraints, let's analyze if the feasible region is unbounded in the direction of decreasing $$Z = 5x - 2y$$. Rewrite the inequalities: $$2x + 3y \leq 12 \quad \text{implies} \quad y \leq \frac{12 - 2x}{3}$$ $$x - y \leq 1 \quad \text{implies} \quad y \geq x - 1$$ $$y \geq 0$$ The feasible region is determined by finding intersections of lines defined by: - $$y = \frac{12 - 2x}{3}$$ - $$y = x - 1$$ Solving for intersection: $$\frac{12 - 2x}{3} = x - 1 \quad \Rightarrow \quad 12 - 2x = 3x - 3$$ $$15 = 5x \quad \Rightarrow \quad x = 3, y = 2$$ Since $$Z=5x-2y$$ will decrease indefinitely as $$x \rightarrow \infty$$, the constraint $$x-y \leq 1$$ must not restrict x from increasing indefinitely. Thus, Z has no minimum if the region allows indefinite increase of x along $$x - y = 1$$, indicating an unbounded feasible region.

Key Concept: Basic Definition

b) Finding the optimal value of a linear function
[Solution Description] A linear programming problem is primarily concerned with finding the optimal value (maximum or minimum) of a linear function of several variables, known as the objective function.

Key Concept:  Basic Constraints

b) $$ 2x + 3y \geq 6 $$
[Solution Description] The constraints are expressed as inequalities or equations that restrict the values of variables. Here, the given constraints are $$ x + y \leq 5 $$, $$ 2x + 3y \geq 6 $$, and non-negativity constraints $$ x \geq 0 $$, $$ y \geq 0 $$.

Key Concept: Basic Definition

b) Variables that represent choices available to the decision-maker
[Solution Description] Decision variables represent the choices available to the decision-maker in a linear programming problem. These variables are central to defining and shaping both the objective function and the constraints within such models.