 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says solve the following linear programming problem graphically and show that the minimum of set occurs at more than two points. We are given that we have to maximize z is equal to minus x plus 2y subject to the constraints x is greater than equal to 3, x plus y is greater than equal to 5, x plus 2y is greater than equal to 6, y is greater than equal to 0. Let us now start with the solution. Now we have to maximize z is equal to minus x plus 2y subject to the constraints x is greater than equal to 3, x plus y is greater than equal to 5, x plus 2y is greater than equal to 6, y is greater than equal to 0. Now for drawing the graph and finding the feasible region subject to given constraints first of all we shall draw the line representing the equation x is equal to 3 corresponding to the inequality x is greater than equal to 3. Now we know x is equal to 3 is a line parallel to y axis also x is equal to 3 passes through the point 3 0. So this is a line x is equal to 3. Now we shall draw a line representing the equation x plus y is equal to 5 corresponding to the inequality x plus y is greater than equal to 5. Now we find that the points 0, 5 and point 5, 0 lie on the line x plus y is equal to 5. So the graph of the line can be drawn by plotting the points 0, 5 and 5, 0 and then joining them. Now this line represents x plus y is equal to 5. Now we shall consider the half plane satisfying x plus y is greater than 5 that is the half plane which does not contain origin. How clearly we can see line corresponding to the inequality x plus 2y greater than equal to 6 is x plus 2y is equal to 6. So we shall draw the line representing the equation x plus 2y is equal to 6 on the same graph by plotting the points 0, 3 and 6, 0 which satisfy the equation x plus 2y is equal to 6. Now we will plot the points 0, 3 and 6, 0 and join them to get the line x plus 2y is equal to 6. Now this line represents x plus 2y is equal to 6. This line divides the plane into two half planes. We will consider the half plane satisfying x plus 2y is greater than 6 that is the half plane not containing 0, 0 or we can say the half plane not containing origin. Now next inequality given to us is y is greater than equal to 0. Now this inequality implies that the graph lies above x axis only. Now this shaded portion in the graph satisfies all the given constraints. This shaded region is the feasible region. We know the common region determined by all the constraints including the non-negative constraints. x is greater than equal to 0, y is greater than equal to 0 of a linear programming problem is called the feasible region. Now in this given problem this is the feasible region. Now clearly we know that the feasible region is unbounded from above and it extends to infinity. The corner points of the feasible region on the lower side are 3, 2, 4, 1 and 6, 0. Now let us calculate the value of objective function z is equal to minus x plus 2y at these three corner points at corner point 3, 2 value of z is equal to 1. We know minus 1 multiplied by 3 plus 2 multiplied by 2 is equal to 1. At corner point 4, 1 value of z is equal to minus 2. We know minus 1 multiplied by 4 plus 2 multiplied by 1 is equal to minus 2. Now at corner point 6, 0 value of z is equal to minus 6, minus 1 multiplied by 6 plus 2 multiplied by 0 is equal to minus 6. Now clearly we can see maximum value of z is equal to 1 which occurs at the corner point 3, 2. But we can locate a point say 6, 6 which belongs to the feasible region and value of z at this point is 6. Now we can write but we can locate a point say 6, 6 which belongs to feasible region and value of z at this point 6, 6 is z is equal to minus 1 multiplied by 6 plus 2 multiplied by 6 which is further equal to 6 only. Now we know 6 is greater than 1 or we can say value of z at point 6, 6 is greater than the maximum value of z at a corner point of the feasible region. Now we know if the feasible region is unbounded and m is the maximum value of the objective function at any point of the corner points and a value greater than m belongs to the feasible region then the objective function has no maximum value. Now here 1 is the maximum value of the given function at corner point 3, 2. Also 6 is a value greater than maximum value and it belongs to feasible region. So there is no maximum value of z. So our required answer is z has no maximum value. This completes the session hope you understood the solution take care and keep smiling.