 Hello and welcome to the session. In this session we will discuss the graphical method of solving linear programming problems. The graphical method of solving linear programming problems involves two main steps. The first one being to find out the feasible region, the feasible solution. Now after this our next step is to determine the optimal solution from the feasible region. Now let us define the terms feasible region, feasible solution and the optimal solution. First we have the feasible region. It is the common region determined by all the constraints including the non-negative constraints. So for linear programming problem when we graph all the constraints the common reason determined by all these constraints is the feasible region. The next term is feasible solutions. These are the points within on the boundary of the feasible region. So when we have the feasible region with us, then the points which are on and within the boundary of the feasible region are the feasible solutions. The point outside the feasible region is called an infeasible solution. Next term is the solution in the feasible region that gives the optimal value which is the maximum or the minimum value of the objective function is called an optimal solution. There are many points in the feasible region that satisfy all the constraints. So to find out the point that gives the maximum value or the minimum value of the objective function we would use two theorems. We have the theorem one according to which we have let the feasible region for a linear programming problem let z equal to ax plus vy be the objective function when z that is the objective function has an optimal value that is the maximum or the minimum value where the variables in vy take two constraints described by linear qualities optimal value must occur at a corner point of the feasible region. This corner point is a point in the region which is the intersection of two boundary lines. So in this theorem we have that if we have that r is only feasible region for the linear programming problem and z is the objective function given in the form of ax plus vy then the optimal value of the objective function that is z must occur at the corner point of the feasible region. Now the second theorem is I will be the feasible region for a linear programming problem and let z equal to plus vy be the objective function if the feasible region r is bounded then the objective function z has both maximum or minimum value on r and each of these occur at a corner point the feasible region r. So we are given r is the feasible region and z is the objective function which is in the form of ax plus vy if it is given to us that the feasible region is bounded which means that it can be enclosed within a circle then the objective function z will have both a maximum and a minimum value on the feasible region r and each of these values would occur at the corner point of the feasible region. Let us now consider an example of a linear programming problem find its solution graphically consider a linear programming problem to maximize the profit function given by z equal to 50 x plus 18 y subject to the constraints given as 2 x plus y less than equal to 100 x plus y less than equal to 80 x greater than equal to 0 and y greater than equal to 0. Like this objective function we equation 1 is constraints p 2 3 and 4 now we are supposed to solve the linear programming problem graphically. Now since we know that the graphical method of solving a linear programming problem involves two steps to find out the feasible region and the feasible solution and in the second step we find the optimal solution from the feasible region. So first of all let's find out the feasible region and as we know that feasible region is the common region determined by all the constraints including the non-negative constraints. So for this we have to graph these constraints and the common region which would be determined by these constraints would be our feasible region. Now to graph this inequality first we will graph the equation 2 x plus y equal to 100 this line joining the points a and b is the graph of the equation 2 x plus y equal to 100. Now we will see that if the origin satisfies this inequality 2 so for x equal to 0 and y equal to 0 in the equation 2 we have 2 into 0 plus 0 less than equal to 100 that is 0 is less than equal to 100 and this is true therefore this means that the origin of which coordinates 0 0 lies in the region given by the inequality 2 that is 2 x plus y less than equal to 100. So the region containing the origin and below this line a b is the solution set of the inequality 2 x plus y less than equal to 100. So this region so region r1 is the region of the equation 2 x plus y less than equal to 100. Now next we need to graph the inequality 3 which is x plus y less than equal to 18 for this first of all we graph the equation x plus y equal to 80 so this line joining the points c and d is the graph of the equation x plus y equal to 80 now for x equal to 0 by equal to 0 in equation 3 we have 0 plus 0 is less than equal to 80 that is 0 is less than equal to 80 which is true therefore this means that the origin lies in the region plus y less than equal to 80. So the region below this line cd containing the origin would represent the in equation x plus y less than equal to 80. So this dimension represented by the red color given as r2 is the region x plus y less than equal to 80. Now next we have the non-negative constraints x greater than equal to 0 which would be the y axis and the region on its right hand side and in the same way the constraint y greater than equal to 0 would be x axis and the region above the x axis. Now that we have graphed all the constraints so we can easily find out the feasible region as we know that feasible region is the common region determined by all the constraints. Now let this point of intersection of the two lines x plus y equal to 80 and 2x plus y equal to 100 be a point c so this region odcdo with feasible region. The coordinates of this point of intersection of the two lines that is the point c which would be obtained by solving the equations x plus y equal to 80 and 2x plus y equal to 100 plus y equal to 100 plus y equal to 80 for this we would subtract these two equations this would give us x equal to 20 now to find the value of y we would substitute x equal to 20 in any of these two equations so we have 2 into 20 plus y equal to 100 or you can say that y is equal to 100 minus 40 which is equal to 60. Therefore we have x equal to 20 and y equal to 60 we have the coordinates of point c 60 so we have got all the vertices of the feasible region and we know that these points odcd and the points within this region and also the other points which are on the boundary of this feasible region are the feasible solutions for the problem. Now after we have obtained the feasible region and the feasible solutions for the linear programming problem our next step is to find out the optimal solution for the problem and for this we follow the corner point method in which our first step is to find out the feasible region and we have obtained the feasible region as odcdo and as you can see this feasible region is bounded now in the next step we find out the corner points the feasible region which could be obtained either by inspection or by solving the two equations of the lines intersecting at that point you have the corner points of the feasible region as odcd with their coordinates so the corner points are o with coordinates 0 0 b with coordinates 50 0 c with coordinates 20 60 and b with coordinates 0 80. Now on this term stated we have that the objective function has an optimal value and that value must occur at the corner point of the feasible region as the feasible region r is bounded then this objective function has both maximum and the minimum value and each of these would occur at the corner point of the feasible region but as we are supposed to maximize the profit function z so we will find only the maximum value of the objective function at any of the corner points and that would be our optimal solution now to do this further we will evaluate the objective function z at each of these corner points now our first corner point is point o with coordinates 0 0 and for this point the objective function z is equal to 50 into 0 plus 18 into 0 and this is equal to 0 now the next corner point is the point b with coordinates 50 0 at this the objective function z would be equal to 50 into 50 plus 18 into 0 and this is equal to 20 500 next corner point is the point c with coordinates 20 60 for this the objective function z is equal to 50 into 20 plus 18 into 60 which would be equal to 2080 next corner point is the point b with coordinates 0 80 and for this the objective function z would be equal to 50 into 0 plus 18 into 80 and this is equal to 1440 now as these values of the objective functions calculated at these corner points you can see that 2500 is the maximum value so let us denote this maximum value of the objective function by capital M and this is equal to only 500 this is the maximum value of z and get small m with the minimum value of z which is 0 obtain the maximum value of the objective function z as 2500 and this is at the point b with coordinates 50 0 so we say that the point b with coordinates 50 0 is the solution z equal to 2500 is the value so this is how we solve a linear programming problem graphically to find the feasible region for the problem and the feasible solutions and then using the corner point method we find the optimal solution for the problem this completes the session hope you have understood the graphical method of solving a linear programming problem