 So, Myself Satish Thalange, assistant professor, department of civil engineering, watch and instructor knowledge, in today's session, we are going to see maximization of objective function of the linear programming problem using graphical method. In today's session, the learner will be able to determine the solution of linear programming problem for maximizing of objective function using graphical method. So, the linear programming problem, the linear programming problem is a mathematically modeling technique in which the following terms are involved. First is objective function in which it may be maximized case or minimized case. Second, the n decision variables are present. These decision variables are non-negative variables, which are subjected to the set of constraints expressed by linear equation, which may be equality equation or inequality equation. The object of linear programming is to maximize the profit, benefits and production of the goods. Here, it is minimize the time, loss and the wastage of resource. Finally, its objective is for the planning and making of the decision about resource allocation. Now, the obtained linear programming problem can be solved by following mathematical techniques. First one is graphical method, which is also called as two variable method. Second simplex, third big M method and last is two phase method. The graphical method is suitable for solving the linear program problems having only two decision variables, as it is going to be solved by the graphical method, which is having only two axis, x axis and y axis. For so, it is suitable to the problem, which is having only two decision variables. And it is suitable to find out the feasible region, which is satisfying almost all the constraints, which is helping us to find out the highest or the lowest point, according to the case to define the optimum solution for the problem. These are the steps, which are involved for solving of the graphical method. Now, let us take the example and start to solve step by step the particular LPP problem. This is a problem, in which we are observing that the company is manufacturing the chair and the table. Here the company is defining the profit, that is unit profit for the table and the chair, for the table it is 60 and for the chair it is 80. And it is also defining the constraints, that is ability of the raw material, that is wood, which is only 300 board feet. And ability of the labour time in hours, that is 110 labour hour ability. Now let us convert this particular problem in the LPP form. The first step is to define the objective function. As we have seen that in this particular question, the company is interested to maximise its profit. So here we have to define the object of the particular companies to maximise its unit profit. Here x 1 and x 2 are the two non-negative decision variables of table and chair. The maximise case z is equal to 60 x 1 plus 80 x 2. Second step we have to define the constraints. There are the two constraints, as we have observed in this table, wood and the labour are the two constraints. So let us convert this particular constraints in the equation. As the company is giving the restriction of particular raw materials, availability and the labour hour ability. Now this particular equation will be inequality equation. This is the equation which is obtained for the wood, availability and labour hours ability. 30 x 1 plus 20 x 2 less than or equal to 200, 5 x 1 plus 10 x 2 less than or equal to 110. And this obtained, we have to give the restriction to the particular decision variables which is greater than or equal to 0, which is most important. Now with the help of the above formulated equations, we have to plot the graph. Here the blue colour line is representing the first constraint that is of wood constraint and second brown colour is representing the second constraint that is labour hour ability. One of the example I have shown to get the values of x 1 and x 2 for the first constraint. Here we are going to place first x 2 value as 0 and we are going to obtain the x 1 value that is a and particularly we are going to place x 1 as 0 and we are going to find out x 2 value. Here these are the points of the first constraint that is of wood constraints. With the help of this a and b, we are going to plot the line of the wood constraint. Similarly, we have to carry for the second constraint. Now let us see the feasible region. Here once we have obtained the feasible 1, 2 and 3 that is area x, y and z. The area x is representing the region which is satisfying only one constraint that is of wood and the area y is the region which is satisfying both the constraints that is of wood and labour hours ability and area z is the region which is satisfying only the second constraint that is labour hour ability. After this three, area y is a region which is satisfying all the constraints available in the particular problem that is of wood constraint and labour constraints. So here the arrows are towards the origin because we have the limitations means there is not more than ability of 300 board feet of raw material and 110 labour hour with us. So that is why the arrows are towards the particular origin. Now once we have finalized this particular feasible region, this is a region which is satisfying the both the constraints. So we have to find out a such a point in this region or on the boundary of the region which is giving the optimum solution for the particular problem. Now according to the maximize case we have to see the corner points of the particular graphical methods of feasible region. So particularly O is the origin, A is the one more corner, P is the intersection point and D is the one more corner of the particular feasible region. Let us take this coordinates values as I have shown in this particular table. These are the points A, D, P and O and its coordinates values and these are the objectives functions equation that is equal to 60 x 1 plus 80 x 2. Why as I place the values of point A in objective function I am getting the value of z for similarly we have to find out for the A, D, P and O. When we observe this table P is a point which is getting the maximum value that is 960 and this is a P is a point here we see this is a point which is satisfying both the constraints because this is a point where both the constraints are satisfying the above problem for maximizing the profit. Once we get the point we get the value of particular problem saying that the company should manufacture 4 number of tables and 9 number of chairs which is helping to maximize its profit by the units of tables and chairs. Here the profit is z is equal to 960. Now let us see the advantages and disadvantages. The advantages what it is a graphical representation that is a picture representation it is faster to solve and understand here we can visually see the feasible region and make a decision confidently and finally it is more simple as compared to the other techniques. Disadvantages are here the LPP having only two variables can be solved by this method. Here we cannot solve the particular problems with a smaller interval of perimeter changing. Here it is not suitable to make a decision with the help of the graphical method. These are the questions let us select the correct answer for this. These are the answers for the above questions. These are the references for today's session. Thank you.