 How are you welcome to the session? I'm Shashi and I'm going to help you with the following question. Question says, show that the minimum of set occurs at more than two points. Minimize and maximize z is equal to x plus 2y subject to x plus 2y greater than equal to 100, 2x minus y less than equal to 0, 2x plus y less than equal to 200, x greater than equal to 0 and y greater than equal to 0. Let us now start with the solution. Now we have to minimize and maximize z is equal to x plus 2y subject to x plus 2y greater than equal to 100, 2x minus y less than equal to 0, 2x plus y less than equal to 200, x greater than equal to 0 and y greater than equal to 0. Let us name these inequalities as 1, 2, 3, 4 and 5. Now for drawing the graph and finding the feasible region subject to given constraints, first of all we shall draw a line representing x plus 2y is equal to 100 corresponding to this inequality. Now points 0, 50 and 100, 0 lie on the line x plus 2y is equal to 100. Now we will draw this line by plotting these two points on the graph and joining them. This point represents 0, 50 and this point represents 100, 0. Joining these two points, we get the line x plus 2y is equal to 100. Now this line divides the plane into two half planes. Now the plane that does not contains the origin that is the plane above this line represents x plus 2y greater than 100. So we will consider this plane. Now we will draw a line 2x minus y equal to 0 corresponding to this inequality. Now points 0, 0 and 50, 100 lie on the line 2x minus y is equal to 0. Now we will plot these two points on the same graph to draw this line. Clearly we can see this point represents 0, 0 and this point represents 50, 0. Joining these two points, we get the line 2x minus y is equal to 0. Here, clearly we can see this line intersects previous line that is x plus 2y is equal to 100 at point p having coordinates 20, 30. Now again clearly we can see this line divides the plane into two half planes. Now this plane above this line satisfies 2x minus y less than 0. Or we can say this plane represents 2x minus y is less than 0. So we will consider this plane. Now we will draw a line 2x plus y is equal to 200 corresponding to this inequality. Now points 0, 200 and 100, 0 lie on the line 2x plus y is equal to 200. Now we will plot these two points on the same graph and draw this line by joining them. Now this point represents 0, 200 and this point represents 100, 0. Now joining these two points, we get this line and this line represents 2x plus y is equal to 200. Now again this line divides the plane into two half planes. One above this line and one below this line. Now half plane below this line represents 2x plus y less than 200. So we will consider this plane. We are also given that x is greater than equal to 0 and y is also greater than equal to 0. This implies that the graph lies in the first quadrant only. You know this line represents y is equal to 0 and this half plane represents y greater than 0. Similarly, this line represents x is equal to 0 and this half plane represents x is greater than 0. So we get the graph lies in first quadrant. Now we will consider all the half planes satisfying the given constraints and shape the common region determined by all the constraints. We get this convex polygon. Now let us name this point as a this point as c and this point as d. So corner points of this polygon are a, p, c and d. This convex polygon represents the feasible region. Let us recall that the common region determined by all the constraints including the non-nibbative constraints x is greater than equal to 0 and y is greater than equal to 0 of a linear programming problem is called the feasible region. So this region is the feasible region. Now according to corner point method the maximum or minimum value of a linear objective function over a convex polygon or we can say feasible region determined by all constraints occurs at some vertex of the polygon. Now we can write the shaded region is the feasible region a, p, c, d determined by system of constraints 1 to 5 which is bounded. Now the coordinates of corner points a, p, c and d are 0, 50, 20, 40, 50, 100 and 0, 200. Now we will evaluate z is equal to x plus 2y at these points. Now substituting x is equal to 0 and y is equal to 50 in this expression we get z is equal to 1 multiplied by 0 plus 2 multiplied by 50 which is further equal to 100. So we can write z is equal to 100 at 0, 50. Similarly z is equal to 100 at 20, 30 substituting 20 for x and 30 for y. In this expression we get z is equal to 100. Now we will find the value of z at 50, 100. So we can write 1 multiplied by 50 plus 2 multiplied by 100 is equal to z. Now simplifying this expression we get z is equal to 250. So we can write z is equal to 250 at 50, 100. Similarly z is equal to 400 at 0, 200. Now clearly we can see maximum value of z is 400 that occurs at 0, 200 and minimum value of z is 100 that occurs at 2 points. They are 0, 50 and 20, 50. Now we can write hence the maximum value of z is equal to 400 which occurs at 0, 200 and minimum value of z is equal to 100 which occurs at 2 points. These 2 points are 0, 50 and 20, 30. Now we know that if the maximum or minimum value occurs at 2 points of the feasible region then it occurs at all the points of the segment joining those 2 points. So minimum value of z is equal to 100 occurs at all the points on the line segment joining these 2 points. Now this is the line which joins these 2 points. So minimum value of z occurs at all the points on the line segment joining these 2 points. Now we can write minimum value of z is equal to 100 occurs at all the points on the line segment joining the points 0, 50 and 20, 50. So this is a very quiet answer. This completes the session. Hope you understood the solution. Take care and have a nice day.