 Hello and welcome to the session. In this session we discuss the graphical method of solving linear programming problems for unbounded feasible region. A feasible region is unbounded means that it extends indefinitely in any direction. We know that in solving a linear programming problem we have two steps in which first we find the feasible region and the feasible solutions then in the next step we find out the optimal solution for that problem. The feasible region is unbounded then maximum a minimum value to objective function may not exist. Moreover it exists that is the maximum or the minimum value of the objective function exist then it must occur at a corner point of the feasible region. Let us consider an example of a linear programming problem in which the feasible region is unbounded. Consider the linear programming problem 6x plus 10y subject to constraints given as s2y greater than equal to 12, 3x plus y greater than equal to 8 and the non-negative constraints x greater than equal to 0, y greater than equal to 0. We shall use the corner point method to solve this problem that involves the following steps in which the first one is to find the feasible region. We know that the feasible region is the common region determined by these constraints. So to find the feasible region we need to graph these inequalities. First of all let's name this objective function as 1, this constraint 2, this constraint 3 and this constraint 4. Now let us start with this first inequality that is this constraint. We first graph the inequality for this first we graph the equation 2x plus 2y equal to 12 or you can say x plus y equal to 6. This line joining the points A and B is the graph of the equation 2x plus 2y is equal to 12. Let's see if the origin satisfies this inequality. So for x equal to 0, y equal to 0 in inequality 2 we have 2 into 0 plus 2 into 0 is greater than equal to 12 which means that 0 is greater than equal to 12 and this is not true and this means that the origin does not lie in the region plus 2y greater than equal to 12. So this shaded region in yellow that this vr1 represents the region 2x plus 2y greater than equal to 3. Now in the same way to graph inequality 3 we first of all graph the equation 3x plus y is equal to 8. This line joining the points C and D represents the graph of the equation 3x plus y equal to 8. Let's see if the origin satisfies the inequality. So for x equal to 0 and y equal to 0 in inequality 3 that is this inequality we have 3 into 0 0 greater than equal to 8 that is 0 greater than equal to 8. This is not true that the origin does not lie in the region 3x plus y greater than equal to 8. So this region in red say r2 represents the region 3x plus y greater than equal to 8. Now the remaining non-negative constraints x greater than equal to 0 and y greater than equal to 0 we know that x greater than equal to 0 is the y-axis that is this and the region to the right of the y-axis and equal to 0 is x-axis and the region above x-axis. Now that we have graph all the constraints so we can easily find out the feasible region. This feasible region would be the common region determined by these constraints. This region determined by these arrows is the feasible region and as you can see that the feasible region is unbounded. So thus we have obtained the region 3db feasible region is finding the feasible region is done. In the next step we find out the corner point cd. c has coordinates 0, 8, d has coordinates 1, 5 which is the point of intersection of the two lines has coordinates. So there are three corner points of the unbounded feasible region point c which has coordinates 0, 8, d which has coordinates 1, 5 and d which has coordinates 6, 0. The function given by z equal to 10y at these corner points corner point c which coordinates 0, 8. Let's find out the value of the subjective function z at this corner point z would be equal to 6 into 0 plus 10 into 8 and so this is equal to 80. Then the next corner point is the point d which coordinates 1, 5. z is equal to 6 into 1 plus 10 into 5 which would be equal to 56. Next we have the corner point d which coordinates 6, 0. So the value of the objective function z is equal to 6 into 6 plus 10 into 0 and this is equal to 36. Here the greatest value of the objective function z is 18. Let capital M be equal to 18 and the smallest value of the objective function z is 36. Let small m be equal to 36. First to minimize the objective function so this means we have to find the minimum value of the objective function z to the region equal to let it be small m equal to 36 not be the minimum value of the objective function z. We graph the inequality 10y less than 36. For this first of all we graph the equation 66 plus 10y equal to 36. This line joining the points b and e is the graph of the equation 10y is equal to 36. Let this inequality be in equation 5. So to graph this inequality we first check that if the origin satisfies this equation or not. So for x equal to 0, y equal to 0 and 5 we have 6 into 0 plus 10 into 0 is less than 36 which means that 0 is less than 36 to 2. Thus we have the origin lies in the region 6x plus 10y is less than 36. So this region below this line bd the region of the equation is this 10y less than 36 and half determined by the equation 6x less than by less than 36 does not have a common point with the feasible region therefore we can say that 36 is the minimum value of the objective function. Thus we have the region 6x plus 10y less than 36 has no point common with the feasible region therefore 36 is the minimum value the objective function z equal to 6x plus 10y. The region determined by this inequality would have any point common with the feasible region then the objective function would not have any minimum value. If suppose you have an objective function z in the form of x plus dy is the maximum value of z determined at some corner point of the feasible region then to check if this m graph the inequality dy greater than dy greater than m no point in common feasible region maximum value of the objective function z and otherwise you have determined by a x plus dy greater than m has points in common with the feasible region then the objective function z has a maximum value. So this is the method of solving a linear programming problem graphically when the feasible region is unbounded. This completes the session. Hope you have understood the method.