 Hi and welcome to the session. Let us discuss the following question. Question says, show that minimum asset occurs at more than two points. Maximize z is equal to x plus y subject to x minus y less than equal to minus 1 minus x plus y less than equal to 0, x greater than equal to 0, y greater than equal to 0. Let us now start with the solution. Now we have to maximize z is equal to x plus y subject to constraints x minus y less than equal to minus 1, minus x plus y less than equal to 0, x greater than equal to 0 and y greater than equal to 0. Now to draw the graph and find the feasible region subject to given constraints we shall first draw line representing the equation x minus y is equal to minus 1 corresponding to inequality x minus y less than equal to minus 1. Now we find that these three points that is 0, 1, 2, 3, minus 1, 0 lie on line x minus y is equal to minus 1. So we can draw this line by plotting these three points on the graph. Now this point represents point 2, 3, this point represents 0, 1 and this point represents point minus 1, 0. Now joining these three points we get the line x minus y is equal to minus 1. Clearly we can see this line divides the plane into two half planes. Now the half plane which does not contain the origin 0, 0 is the plane which satisfies the inequality x minus y is less than minus 1. Now we will draw a line minus x plus y is equal to 0 corresponding to this inequality. Clearly we can see points 0, 0 and 1, 1 lie on this line. So we can write 0, 0 and 1, 1 are points lying on minus x plus y is equal to 0. Now this point represents the point 1, 1 and this point represents point 0, 0. Now joining these two points we get the line minus x plus y is equal to 0. Clearly we can see this line divides the plane into two half planes. One is above this line and other below this line. Now the plane which satisfies the inequality minus x plus y is less than 0 is below this line. We are also given that x is greater than equal to 0 and y is greater than equal to 0. This implies that the graph lies in the first quadrant only. Now clearly we can see this shaded region satisfies the inequality x minus y is less than equal to minus 1 and this shaded region satisfies the inequality minus x plus y is less than equal to 0. We also know that given graph lies in the first quadrant. So both the shaded regions are lying in first quadrant only. Now clearly we can see there is no region common to both of these half planes. So there is no feasible region satisfying all the given constraints. We know that if there is no feasible region there is no maximum or minimum value of a linear objective function. So given objective function that is z is equal to x plus y has no maximum value. Now we can write there is no feasible region in the graph. Therefore there is no maximum value of z. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.