 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says solve the following linear programming problem graphically Maximize z is equal to 5x plus 3y subject to 3x plus 5y less than equal to 15 5x plus 2y less than equal to 10 x greater than equal to 0 and y greater than equal to 0 So let's start the solution now according to the given problem. We have to maximize z is equal to 5x plus 3y Let us give this as number one Subject to the constraints 3x Plus 5y less than equal to 15 5x Plus 2y less than equal to 10 x greater than equal to 0 and y greater than equal to 0 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 Representing 2 is that is corresponding to the inequality 3x plus 5y less than equal to 15 is 3x plus 5y is equal to 15 So here we observe that the points 0 3 And 5 0 lie on the line 3x plus 5y is equal to 15 So the graph of the line 3x plus 5y is equal to 15 can be drawn by plotting points 0 3 and 5 0 and then joining them So let us take a as a point with coordinates 0 3 and b as a point with coordinates 5 0 So a b is the line representing the equation 3x plus 5y is equal to 15 now the line a b divides the plane into two half planes Now clearly the origin does not lie on the line 3x plus 5y is equal to 15 It satisfy the inequality 3x plus 5y less than 15 So the closed half plane containing the origin is the graph of 2 Again the line corresponding to the inequality 5x plus 2y less than equal to 10 is 5x plus 2y is equal to 10 So we will draw the line representing the equation 5x plus 2y is equal to 10 on the same graph Clearly the points 0 5 and 2 0 Satisfy the equation 5x plus 2y is equal to 10 So let us take c as a point 0 5 and d as a point 2 0 So the line cd represent the equation 5x plus 2y Is equal to 10 Now this line divides the plane into two half planes So we will consider the half plane satisfying the inequality 5x plus 2y less than 10 So the closed half plane containing the origin is the graph of 3 Also x greater than equal to 0 y greater than equal to 0 Implies that the graph flies in the first quadrant only So OAPD is the feasible region According to the given constraints Now here we observe that the feasible region OAPD is bounded So we will use corner point method to determine the maximum value of safe now according to the corner point method the corner points of the feasible regions are Fine either by inspection or by solving the two equations of the lines intersecting at that point now on solving 3x plus 5y is equal to 15 and 5x plus 2y is equal to 10 we get x is equal to 20 by 19 and Y is equal to 45 over 19 Hence the coordinates of the point P are 20 over 19 and 45 over 19 Now we will evaluate Z at each corner point So we have the coordinates of of the corner points a P and D as 0 0 0 3 20 over 19 45 over 19 and 2 0 respectively at the point O That is at the origin Z is equal to 5 into 0 plus 3 into 0 which is equal to 0 Now at the point a With coordinates 0 3 Z is equal to 5 into 0 plus 3 into 3 which is equal to 9 Again at the point P with coordinates 20 over 19 and 45 over 19 Z is equal to 5 into 20 over 19 plus 3 into 45 over 19 which is equal to 100 over 19 plus 135 over 19 and this is equal to 235 over 19 now at the point D With coordinates 2 0 Z is equal to 5 into 2 plus 3 into 0 which is equal to 10 hence the maximum value of Z equal to 235 over 19 which occurs then X is equal to 20 over 19 and Y is equal to 45 over 19 Hence the answer for the above question is maximum Z is equal to 235 over 19 at the point 20 over 19 45 over 19 So this completes our session. I hope the solution is clear to you Bye and have a nice day