Linear Programming is a mathematical optimization technique used to maximize or minimize a linear objective function subject to linear constraints. It is widely applied in areas such as production planning, diet formulation, transportation, scheduling, and labour allocation. Option B is correct because determining non-linear moving averages of stocks belongs to financial and statistical analysis rather than Linear Programming. Option A is an important application of LPP in diet planning, where the objective is to minimize cost while satisfying nutritional requirements. Option C is a standard application of LPP for maximizing profit in manufacturing. Option D is also a common application, where available labour is allocated efficiently under given constraints.

Optimization in Linear Programming refers to finding the maximum or minimum value of a linear objective function while satisfying a set of linear constraints and non-negative restrictions. The objective may be maximizing profit or production, or minimizing cost or time. Option C is correct because it accurately defines optimization in LPP. Option A is incorrect because plotting variables alone does not perform optimization. Option B is incorrect because decision variables are not required to be zero. Option D is incorrect because constraints limit the solution rather than expanding infinitely.

Constraints in a Linear Programming Problem represent the limitations on available resources. They are expressed as linear equations or linear inequalities and, together with the non-negative restrictions, define the feasible region. Option D is correct because it is the incorrect statement. LPP uses only linear relationships; quadratic equations are not allowed. Option A correctly states the purpose of constraints. Option B correctly describes their mathematical form. Option C correctly identifies non-negative restrictions as part of the LPP formulation.

Each term has a standard mathematical representation in Linear Programming. Objective Function → Z = ax + by Linear Constraint → 3x + 5y ≤ 15 Non-negative Restriction → x ≥ 0, y ≥ 0 Arbitrary Constants → The coefficients a and b Thus, the correct matching is: 1 → a 2 → b 3 → c 4 → d Hence, Option A is correct.

In a Linear Programming Problem, the objective function represents a quantity such as profit, cost, or production and must always be a linear function of the decision variables. A typical objective function is: Z = ax + by where a and b are constants. The Assertion is true because the objective function must satisfy the assumption of linearity. The Reason is false because linear inequalities are first-degree expressions. Quadratic expressions involve variables raised to power 2, which are not allowed in Linear Programming. Therefore, Option B is the correct answer.

The phrase "at most 20 items" means the total quantity cannot exceed 20. If x and y represent the quantities of two items, then the storage condition is: x + y ≤ 20 This inequality allows the total number of items to be exactly 20 or any value less than 20. Option C correctly represents the storage constraint. Option A means at least 20 items, which contradicts the statement. Option B allows only exactly 20 items. Option D is meaningless because quantities of items cannot be negative.

Evaluate Z = 5x + 3y at each corner point. At (0,0) Z = 5(0) + 3(0) = 0 At (4,0) Z = 5(4) + 3(0) = 20 At (0,4) Z = 5(0) + 3(4) = 12 The maximum value among 0, 20, and 12 is 20. Therefore, Option D is the correct answer.

Evaluate the objective function Z = 3x + 5y at each corner point. At (3,0) Z = 3(3) + 5(0) = 9 At (1.5,0.5) Z = 3(1.5) + 5(0.5) = 4.5 + 2.5 = 7 At (0,2) Z = 3(0) + 5(2) = 10 Comparing the values: 9 7 10 The smallest value is 7. Therefore, Option A is the correct answer.

In a diet problem, the objective is to prepare a diet that satisfies the minimum nutritional requirements while keeping the total cost as low as possible. Thus, the objective function is to minimize the total cost, whereas the nutrient requirements form the constraints. Option B correctly states the objective. Option A is incorrect because the goal is not to maximize nutrients but to satisfy the required minimum. Option C is incorrect because the number of food items is not the optimization objective. Option D is incorrect because reducing nutrients to zero would violate the nutritional constraints.

In a transportation problem, goods are shipped from sources to destinations while minimizing transportation cost. The solution must satisfy the available supply at each source and the required demand at each destination. Therefore, the constraints are based on supply and demand requirements. Option C correctly identifies the constraints. Option A is incorrect because fuel cost may influence the objective function, not the constraints. Option B is incorrect because the number of trucks is not the standard constraint in the transportation model. Option D is incorrect because vehicle speed is not a constraint in the basic transportation problem.

Each statement is verified separately. (i) For the equations: x + y = 4 x + 2y = 5 Subtracting the first equation from the second gives: y = 1 Substituting into the first equation: x = 3 Hence, the intersection point is (3, 1). (ii) For the equations: 5x + 2y = 10 3x + 5y = 15 Solving simultaneously gives: x = 20/19 y = 45/19 Thus, the given intersection point is correct. (iii) The inequality x − y ≤ −1 represents one half-plane, while x ≥ y represents another. These together do not produce a bounded feasible region because the solution set, if it exists, extends infinitely rather than forming a closed polygon. Therefore, all three statements are correct. Hence, Option D is the correct answer.

In the graphical method, the objective function is represented by the line: ax + by = M As the value of M increases for a maximization problem, the line shifts parallel to itself. If the half-plane represented by ax + by > M has no common point with the feasible region, then M is the largest attainable value of the objective function. Therefore, M represents the maximum value of Z. Hence, Option A is correct.

In an assignment problem, decision variables represent assignments of employees to tasks. The constraints ensure that the available workforce and working hours are not exceeded while completing the assigned work. Therefore, restrictions generally involve the number of employees and their available work-hours. Option B correctly represents these constraints. Option A refers to the objective function rather than the constraints. Option C relates to the graphical method, not assignment constraints. Option D incorrectly refers to maximizing cost instead of minimizing or optimizing it.

The Iso-profit (or Iso-cost) Method follows these steps: 1. Identify the feasible region and its corner points. 2. Assign a convenient value to Z and draw the objective function line. 3. Draw parallel objective function lines. 4. Move the line parallel until the farthest line still touches the feasible region. Thus, the correct sequence is: 2 → 4 → 3 → 1 Therefore, Option C is correct. Option A begins by selecting the optimum before constructing the graph. Option B draws parallel lines before the initial objective function line. Option D follows the reverse of the required procedure.

The non-negative restrictions in Linear Programming are: x ≥ 0 y ≥ 0 These conditions ensure that both decision variables are either positive or zero. Hence, every feasible solution must lie in the first quadrant of the Cartesian plane. Therefore, Option D is the correct answer. Option A contains negative y-values. Option B contains negative x-values. Option C contains both x and y negative.

The first constraint is: x − y ≤ −1 which can be written as: x ≤ y − 1 The second constraint is: x ≥ y Together they require: x ≤ y − 1 and x ≥ y These conditions cannot be satisfied simultaneously because y − 1 < y. Hence, there is no common solution, and the feasible region is empty. Therefore, Option A is correct.

A feasible solution is any solution that satisfies all the given constraints, including the non-negative restrictions. Every optimal solution must first be a feasible solution. The Assertion is true because it correctly defines a feasible solution. The Reason is false because the optimal solution never lies outside the feasible region. In graphical LPPs, the optimal solution occurs at one of the corner points or along an edge of the feasible region (when multiple optimal solutions exist). Therefore, Option B is the correct answer.

An unbounded feasible region extends infinitely in one or more directions. However, this does not imply that the objective function has no optimal value. In some cases, the objective function reaches a finite maximum or minimum. In other cases, it can increase or decrease indefinitely without attaining an optimum. Hence, the optimal value may or may not exist. Therefore, Option C is the correct answer.

The graphical method of solving Linear Programming Problems is applicable only when there are two decision variables, since the feasible region can then be represented on a two-dimensional graph. Option D is the incorrect statement because graphical methods cannot practically solve LPPs involving three or more decision variables. Option A is correct because linear inequalities are plotted on the coordinate plane. Option B is correct because corner points are obtained from the intersections of constraint boundaries. Option C is correct because the graphical method clearly shows the feasible region. Therefore, Option D is the correct answer.

Given: Z = 22x + 18y Substitute: x = 10 y = 10 Therefore, Z = 22(10) + 18(10) = 220 + 180 = 400 Hence, the value of the objective function is 400. Therefore, Option A is the correct answer.