Structure of linear programming model - class-X

$x \le 4$ and $x \le -4$

Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

The above question asks for the impact of change in constraints on the Linear programming problem. In this scenario, when there is a change in constraint, the solution will change definitely. Whether the solution exists or not, we can only find once the problem is re-evaluated. 

The objective function represents the total profit. If x is quintals of rice (profit 40) and y is quintals of wheat (profit 25), the total profit is 40x + 25y.

the feasibe solution of a inear programming probem(LP) is a solution that must satisfy all of the problem's constraints simultaniously

In linear programming, the set of all points satisfying the constraints is known as the feasible region.

By evaluating the objective function at the corner points of the feasible region defined by the given linear constraints, we find the maximum value. The corner points include the origin, intersection axes points, and the simultaneous solution of the constraint boundary lines, yielding a maximum value of 440.

First $km$ fare =$Rs.10$

The problem associated with LLP is single objective.

Linear programming is an extremely powerful tool for addressing a wide range of applied optimization problems. A short list of application areas is resource allocation, production scheduling, warehousing, layout, transportation scheduling, facility location, flight crew scheduling, portfolio optimization, parameter estimation.

So ,linear programming is used to subset mathematical programming.

In linear programming, $\text{transportation model}$ are applied to problems related to the study of efficient transportation routes.

In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes. i.e., how effectively the available resources are transported to different destinations with minimum cost.

Given constraints, $4x+3y\leq 480$ and $2x+3y\leq 360$

Consider point $(10,120)$ , It doesnt satisfy the inequality $2X+3Y \le 360$

A linear programming problem is said to have unbounded solution if it has infinite number of solutions. I.e., the problem has been formulated improperly

The first step in formulating an linear programming problem is to understand the managerial problem being faced i.e., determine the quantities that are needed to solve the problem.

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints. Then the problem becomes convex and has a single optimum(maximum or minimum) solution. 

if one of the problems(primal, dual) is infeasible then the other problem is infeasible. Hence, the option D is the false statement.

The number of variables in dual is equal to the number of constraints in the primal and the  number of variables in primal is equal to the number of constraints in the dual.

A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.

Option D is correct.

In order to write LPP, it is not necessary that it has at least as many constraints as the number of variables.

there is no limit on constraints allowed in linear programming.

so the number of constraints is unlimited.

For linear programming, the constraints must be linear.

From the primal-dual relationship,

To write the dual, then all the primal variables must be non-negative.

Formulation of L.P Model
Let ${x} _{1},{x} _{2}$ and ${x} _{3}$ denote the number of units of products $A,B$ and $C$ to be manufactured .
Objective is to maximize the profit.
i.e., maximize $Z=90{x} _{1}+40{x} _{2}+30{x} _{3}$
Constraints can be formulated as follows:
For raw material $P, 6{x} _{1}+5{x} _{2}+2{x} _{3}\le5,000$
For raw material $Q, 4{x} _{1}+7{x} _{2}+3{x} _{3}\le6,000$
Product $B$ requires $\frac{1}{2}$ and product $C$ requires ${\left(\frac{1}{3}\right)}^{rd}$ the time required for product $A.$
Then $\frac{t}{2}$ and $\frac{t}{3}$ are the times in hours to produce $B$ and $C$ and since $1,600$ units of $A$ will need time $1,600t$ hours, we get the constraint,
$t{x} _{1}+\frac{t}{2}{x} _{2}+\frac{t}{3}{x} _{3}\le 1,600t$ or 
${x} _{1}+\frac{{x} _{2}}{2}+\frac{{x} _{3}}{3}\le1,600$ or
$6{x} _{1}+3{x} _{2}+2{x} _{3}\le9,600$
Market demand requires
${x} _{1}\ge300, {x} _{2}\ge250,$ and ${x} _{3}\ge200$ 
Finally, since products $A,B$ and $C$ are to be produced in the ratio $3:4:5,$
${x} _{1}:{x} _{2}:{x} _{3}::3:4:5$
or $\frac{{x} _{1}}{3}=\frac{{x} _{2}}{4},$
and $\frac{{x} _{2}}{4}=\frac{{x} _{3}}{5}.$
Thus, there are two additional constraints
$4{x} _{1}-3{x} _{2}=0$ and $5{x} _{2}-4{x} _{3}=0$ where ${x} _{1},{x} _{2},{x} _{3}\ge0$