 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 3x plus 4y subject to the constraints, x plus y is less than equal to 4, x greater than equal to 0 and y greater than equal to 0. Now let's start the solution. Now according to the given problem we have to maximize z is equal to 3x plus 4y. Now a linear function z is equal to ax plus by where a and b are constants which has to be maximized or minimized is called a linear objective function. So in this problem z is equal to 3x plus 4y is our objective function. So let us give this as number one. Now here x and y are called the decision variables. Now the linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. Now we have to maximize z is equal to 3x plus 4y subject to the constraints, x plus y less than equal to 4, x greater than equal to 0 and y greater than equal to 0. Let us give this as number two and this as number three. Now we will draw the graph and find the feasible region subject to these given constraints. Now the equation of line representing 2 that is representing the inequality x plus y less than equal to 4 is x plus y is equal to 4. So we will first draw the line representing the equation x plus y is equal to 4. Now clearly the points 0, 4 and 4, 0 lie on the line x plus y is equal to 4. Therefore the graph of the line can be drawn by plotting points 0, 4 and 4, 0 and then joining them. Now this is a point 0, 4. Let us take this point as a. So a has coordinates 0, 4 and let us take b as 4, 0. So this is a line a, b representing the equation x plus y is equal to 4. Clearly the origin does not lie on this line. It lies on the half plane of 2 that is it satisfies the inequality x plus y less than equal to 4. Also x greater than equal to 0 and y greater than equal to 0 implies the graph lies in the first quadrant only. Now the common region determined by all the constraints including non-negative constraints x greater than equal to 0 and y greater than equal to 0 of a linear programming problem is called the feasible region for the problem. So here the shaded portion in this graph is the feasible region satisfying all the given constraints. Now again a feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle. So here the feasible region OAB is bounded. Now there is a theorem which states let R be the feasible region for a linear programming problem and let z is equal to Ax plus By be the objective function. If R is bounded then the objective function z has both a maximum and a minimum value on R and each of these occurs at a corner point of R. So here the coordinates of the corner points O, A and B are 0 0, 0 4, 4 0 respectively. Now we will evaluate z at each corner point. So at the point O with coordinate 0 0, z is equal to 3 into 0 plus 4 into 0 which is equal to 0. Now at the point A with coordinates 0 4, z is equal to 3 into 0 plus 4 into 4 which is equal to 16 and at B with coordinates 4 0, z is equal to 3 into 4 plus 4 into 0 which is equal to 12. Therefore maximum value of z is 16 which occurs when x is equal to 0 and y is equal to 4 that is at the point 0 4. Hence the answer for the above question is maximum z is equal to 16 at 0 4. So this completes our session. I hope the solution is clear to you. Bye and have a nice day.