 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says Solve the following linear programming problem graphically minimize z is equal to minus 3x plus 4y subject to x plus 2y less than equal to 8 3x plus 2y less than equal to 10 x greater than equal to 0 y greater than equal to 0 So let's start the solution According to the given problem. We have to minimize z is equal to minus 3x plus 4y subject to the constraints x plus 2y less than equal to 8 3x plus 2y less than equal to 12 x greater than equal to 0 and y greater than equal to 0 So z is equal to minus 3x plus 4y is our objective function. Let us give this as number one Let us give numbers to these constraints as 2 3 and 4 Now we will draw the graph and find the feasible region subject to these given constraints Now the equation of the line corresponding to 2 is that is corresponding to the inequality x plus 2y is less than equal to a is x plus 2y is equal to a So first we will draw the graph of the line representing the equation x plus 2y is equal to 8 Now clearly the points 0 4 and 8 0 lie on the line x plus 2y is equal to 8 Therefore the graph of the line can be drawn by plotting points 0 4 and 8 0 and then joining them. So let us take a as a point 0 4 and b as a point 8 0 So this is the line a b representing The equation x plus 2y is equal to a Now the origin does not lie on this line It lies on the half plane of 2 That is the half plane containing origin satisfy this inequality Again the line corresponding to the inequality 3x plus 2y less than equal to 12 is 3x plus 2y is equal to 12 Now the points 0 6 and 4 0 Satisfy the equation 3x plus 2y is equal to 12 So let us take c as a point 0 6 And d as the point 4 0 So cd is the line Representing the equation 3x plus 2y is equal to 12 So again the line cd divides the plane into two half planes So the closed half plane containing the origin is the graph of 3 Also x greater than equal to 0 and y greater than equal to 0 implies The graph lies in the first quadrant only Again here we observe that the point of intersection of two lines X plus 2y is equal to 8 and 3x plus 2y is equal to 12 Is the point 2 3 Let us take this point as p. So p has coordinates 2 3 So here the shaded region OAPD is the feasible region Determined by the system of constraints 2 to 4 now here the feasible region OAPD is bounded So we will use corner point method to determine the Minimum value of set Now according to the corner point method first we will find out the corner points of the feasible region. So here the corner points of this feasible region are OAPD now the coordinates of the corner points OAPD and DR Now the coordinates of OOR 00 A has coordinates 0 4 p has 2 3 and D has 4 0 04 2 3 4 0 respectively Now we will evaluate Z at each corner point Now Z is equal to minus 3x plus 4y So at 00 that is at origin Z is equal to minus 3 into 0 plus 4 into 0 which is equal to 0 now at A with coordinates 04 Z is equal to minus 3 into 0 plus 4 into 4 which is equal to 16 Now at the point P with coordinates 2 3 Z is equal to minus 3 into 2 plus 4 into 3 Now this is minus 6 plus 12 which is equal to 6 Now at the point D with coordinates 4 0 Z is equal to minus 3 into 4 plus 4 into 0 which is equal to minus 12 hence the minimum value of Z is minus 12 which occurs when X is equal to 4 and Y is equal to 0 Hence the answer for the above question is minimum Z is equal to minus 12 which occur at the point 40 So this completes our session. I hope the solution is clear to you buy and have a nice day