Explanation:

Linear programming (LP)

• Linear programming (LP) in industrial engineering is used for the optimization of our limited resources when there is a number of alternate solutions possible for the problem like material selection.
• The real-life problems can be written in the form of a linear equation by specifying the relation between its variables. 
• Linear programming is used for obtaining the most optimal solution for a problem with given constraints.
• Linear programming requires the creation of inequalities and then graphing those to solve problems.
• Using linear programming requires defined variables and constraints, to find the largest objective function (maximization).
• In some cases, linear programming is instead used for the smallest possible objective function value (minimization).
• Some linear programming can be done manually.
• When the variables and calculations become too complex and require the use of computational software.

CONCEPT:

At the corner point of the feasible reason, we have to check the value of the objective function to get the maximum value.

CALCULATION:

Given:

The feasible region as shown in the figure has the objective function F = 3x - 4y.

Hence, the maximum value of F is 12.

So, the correct answer will be option 3.

Explanation:-

Objective function

• A linear function of two or more variables which has to be maximized or minimized under the given restrictions is called an objective function.
• The variables used in the objective function are called as decision variables.

Constraints:

• These are the restrictions on the variables of an linear programming problem are called as linear constraints.
• The final solution of the objective function must satisfy these constraints.

Additional Information

Other terms related to LPP

Linear constraints

• The linear inequalities or restrictions on the variables of a linear programming problem are called as linear constraints.
• The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.

Optimization problem

• A problem which seeks to maximize or minimize a linear function subject to certain constraints as determined by a set of linear inequalities is called a optimization problem

CONCEPT:

• At the corner point of the feasible reason, we have to check the value of the objective function to get the maximum value.

CALCULATION:

Given:

So, the condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) can be extracted by equating the minimum values of Z for (3, 0) and (1, 1).

⇒ p + q = 3p

⇒ 2p = q

∴ \(\rm p=\frac{q}{2}\)

• So, the correct answer will be option 2.

Explanation:

Maximizing z = x₁ - x₂

Constraints x₁ + x₂ ≤ 10

x₁ ≥ 0

x₂ ≥ 0

x₂ ≤ 5

Corner points are: (0, 0) (5, 0), (5, 5), (0, 10).

Z (0, 0) = 0 - 0 = 0

Z (0, 5) = 0 - 5 = -5

Z (5, 5) = 5 - 5 = 0

Z (10, 0) = 10 - 0 = 10

Maximum Z = Z (10, 0) = 10

Thus, problem has one optimum solution.

If we draw the feasible region by given constraints then 

\(\begin{array}{l} \frac{{{x_1}}}{3} + \frac{{{x_2}}}{1} \ge 1\\ \frac{{{x_1}}}{2} + \frac{{{x_2}}}{2} \ge 1 \end{array}\)

Z @ (3, 0) = 4.5 (optimal)

Z @ (0, 2) = 9 

Z @ (3/2, 1/2) = 4.5 (optimal)

So there are multiple optimal solution for the given minimization problem.

In the figure, there are three constraints. But the feasible region is surrounded by two more lines x – axis and y axis.

So total lines = m+2

Calculation:

Here, for solving this LPP problem, converting all inequality constraints to equality constraints.

x = 4

y = 6

3x + 2y = 18

Now, plotting these on the graph,

The shaded region will be optimum region.

Now, finding value of objective function at all points

Z(0) = 0 + 0 = 0

Z(A) = Z(4, 0) = 6 × 4 + 0 = 24

Z(B) = Z(4, 3) = 6 × 4 + 10 × 3 = 54

Z(C) = Z(2, 6) = 2 × 6 + 6 × 10 = 72

Z(D) = Z(0, 6) = 0 + 6 × 10 = 60

So, the maximum value of objective function is 72.

CONCEPT:

• We will formulate the constraints depending on the availability of the resources.

EXPLANATION:

Given: The number of the products of X and Y types are x  and y.

• For non-negativity constraints,

 x  ≥ 0 and y. ≥ 0 

• You can see from the table that the time taken by machine A to manufacture one X type of product is 1 hr similarly the time taken by machine A to manufacture one Y type of product is 1 hr. The total maximum available time on machine A is 4 hrs,

⇒ 1x + 1y ≤ 4      ....(1)

• Similarly, the time taken by machine B to manufacture one X type of product is 3 hrs, and the time taken by machine B to manufacture one Y type of product is 8 hr. The total maximum available time on machine B is 25 hrs,

⇒  3x + 8y ≤ 24      ...(2) 

• For machine C,

⇒ 10x + 7y ≤ 35      ...(3) 

For Profit Maximization (As the unit prize is mentioned as 150 and 100 Rs per unit for X and Y type)

⇒ Max Z = 5x + 7y

So, the correct answer is option 1

Explanation:

There are special cases in Linear programming problems solved by the graphical method.

i) Infinite or Multi–optimum solution: If the slope of the objective function becomes equal to one of the constraints then infinite no. of solution will exist.

ii) No solution or infeasible solution: In some conditions, a constraint may be inconsistent in such a manner that it is not possible to find a common feasible point that satisfies all the constants then there is no solution to such a problem.

iii) Unbounded solution: In some conditions, constraints are inconsistent in such a manner that the greatest value of objective function goes up to infinite. It simply means that a common feasible region is not bounded by the limit on the constraints.