 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says Solve the following linear programming problem graphically show that the minimum of z occurs at more than two points Minimize and maximize z is equal to 5x plus 10y Subject to x plus 2y less than equal to 120 x plus y greater than equal to 60 x minus 2y greater than equal to 0 xy greater than equal to 0 So let's start the solution Now according to the given problem. We have to minimize I am maximized z is equal to 5x Plus 10y So this is our objective function. Let us give this as number one So we have to minimize and maximize that is equal to 5x plus 10y subject to the constraints plus 2y less than equal to 120 x plus 5 greater than equal to 60 x minus 2y greater than equal to 0 x and y greater than equal to 0 So let us number these constraints as two three four and five So we have to draw the graph and find the feasible region subject to these given constraints now the line Corresponding to the inequality x plus 2y less than equal to 120 is x plus 2y is equal to 120 So we will first draw the line representing the equation x plus 2y is equal to 120 now clearly the points 0, 60 and 120, 0 lie on the line x plus 2y is equal to 120 Therefore the graph of this line can be drawn by plotting points 0, 60 and 120, 0 and then joining them So let us take a as the point 0, 60 and b as the point 120, 0 so a b is the line representing the equation x plus 2y is equal to 120 So a b divides the plane into two half planes Clearly origin does not lie on this line It lies on the half plane of 2 So the closed half plane containing the origin is the graph of 2 Again the equation of the line corresponding to 3 is x plus y is equal to 60 Clearly the points 0, 60 and 60, 0 lie on this line So we will draw the line representing the equation x plus y is equal to 60 on the same graph by plotting the points 0, 60 and 60, 0 and then joining them Now let us take c as the point 60, 0 so a c is the line representing the equation x plus y is equal to 60 Now again a c divides the plane into two half planes We will consider the half plane satisfying x plus y greater than 60 That is the half plane which does not contain the origin Now our next constraint is x minus 2y greater than equal to 0 So the equation of the line corresponding to this inequality is x minus 2y is equal to 0 Clearly the points 0, 0 and 60, 30 Satisfy the equation x minus 2y is equal to 0 So to draw a line x minus 2y is equal to 0 We will plot the points 0, 0 and 60, 30 on the same graph and then join them Now let us take d as the point 60, 30 So od is the line representing the equation x minus 2y is equal to 0 Now the line od divides the plane into two half planes One above the line od and the other below it As the half plane below the line od Satisfy the inequality x minus 2y greater than 0 So we will consider this only Also x greater than equal to 0 y greater than equal to 0 implies that the graph flies in the first quadrant only Now let us take the point of intersection of the line x minus 2y is equal to 0 and x plus y is equal to 60 as the point p Clearly the coordinates of p are 40, 20 Now the shaded region p, c, p, t is the physical region Satisfying all the given constraints Clearly the physical region is bounded So now we will use the corner point method to determine the minimum and maximum value of z Now the coordinates of corner points c, b and dr 40, 20, 60, 0 120, 0 and 60, 30 respectively So according to the corner point method Maximum and minimum values of z will occur at these points And again if two corner points of the physical region are both optimal solutions of the same type That is both produce the same maximum or minimum Then any point on the line segment joining these two points is also an optimal solution of the same type Therefore we will calculate the value of the objective function z is equal to 5x plus 10y at these points Now at the point p with coordinates 40, 20 z is equal to 5 into 40 plus 10 into 20 which is equal to 200 plus 200 and this is equal to 400 Now at the point c with coordinates 60, 0 z is equal to 5 into 60 plus 10 into 0 which is equal to 300 plus 0 which is again equal to 300 Now at the point b with coordinates 120, 0 z is equal to 5 into 120 plus 10 into 0 which is equal to 600 plus 0 and that is again equal to 600 And at the point d whose coordinates are 60, 30 z is equal to 5 into 60 plus 10 into 30 which is again equal to 300 plus 300 which is equal to 600 Hence the minimum value of z equal to 300 which occurs at the point 60, 0 and the maximum value of z is equal to 600 which occurs at 2 that is z has the same value at the point 120, 0 and 60, 30 So this implies the maximum value z is equal to 600 occurs at all the points on the line segment joining the points 120, 0 and 60, 30 Answer for this question is minimum z is equal to 300 at 60, 0 maximum z is equal to 600 at all the points on the line segment joining the points 120, 0 and 60, 30 So this completes our session I hope the solution is clear to you Bye and have a nice day