 Hello and welcome to the session. In this session we will discuss about linear programming. A problem which seeks to maximize or minimize a linear function subject to certain constraints as determined by a set of linear inequalities is called an optimization problem. Linear programming problems are special types of optimization problems. So the basic definition of linear programming problem says it is the one that is concerned with finding the optimal value that is the maximum or minimum of a linear function of several variables subject to the conditions that the variables are non-negative and satisfy a set of linear inequalities. In the term linear programming the term linear implies that all mathematical relations used in the problem are linear relations while more programming refers to the method of determining a particular program or plan of action. Now let's consider a linear programming problem. In this we have that one kind of cake requires 200 grams of flour, 25 grams of fat and the second kind of cake requires 100 grams of flour and 50 grams of fat. We are supposed to find the maximum number of cakes that can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes. Let the number of one kind of cake be equal to x and the number of second kind of cake be equal to y. Now as we know that the first kind of cake requires 200 grams of flour and the second kind of cake requires 100 grams of flour and the total flour that we have is 5 kg or you can say 5000 grams. So we have 200 x plus 100 y less than equal to 5000 or this could also be written as 2 x plus y less than equal to 50. Then we have one kind of cake requires 25 grams of fat and second kind of cake requires 50 grams of fat and total fat that we have is 1 kg or you can say 1000 grams. So we get 25 x plus 50 y less than equal to 1000 or we can say x plus 2 y is less than equal to 40. Now as the number of cakes made are greater than equal to 0 so we have x greater than equal to 0 and y greater than equal to 0. Now the mathematical formulation of this problem is given as maximize z equal to x plus y since we are supposed to find the maximum number of cakes subject to constraints x plus y less than equal to 50, x plus 2 y less than equal to 40, x greater than equal to 0 and y greater than equal to 0. Now let's see what is an objective function. The linear function that is z equal to x plus y in this case which is supposed to be maximized is called a linear objective function. The variables in this objective function that is x and y in this case are called the decision variables. The linear inequalities or equations or restrictions that is these inequalities are called the constraints. The conditions x greater than equal to 0, y greater than equal to 0 are called the non-negative constraints. Now we shall draw the graph and find the feasible regions subject to these constraints. First let's consider the inequality 2x plus y less than equal to 50. To draw the graph for this inequality what we do is, first we draw the graph for the line 2x plus y equal to 50. This is the line 2x plus y equal to 50. This divides the graph into two halves. So we substitute the origin that is 0 0 in the inequality 2x plus y less than equal to 50. Origin satisfies this inequality. So the graph for the inequality 2x plus y less than equal to 50 would be the region below the line 2x plus y equal to 50 that is the region containing the origin. In the same way we will draw the graph for the inequality x plus 2y less than equal to 40. This is the line x plus 2y equal to 40. Now when we substitute the origin that is 0 0 in the inequality x plus 2y less than equal to 40 we see that the origin satisfies this inequality. So the graph for this inequality will be the region below this line x plus 2y equal to 40 that is the region containing the origin. And the graph for x greater than equal to 0 y greater than equal to 0 lies in the first quadrant itself. The point of intersection of these two lines is this point let it be p which coordinates 20 10. The common region determined by all the constraints including the 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 or the solution region for the problem. So this shaded portion, this yellow shaded portion is the feasible region for this linear programming problem. The region other than the feasible region is called the infeasible region and the points within and on the boundary of the feasible region represent the feasible solutions of the constraints like the point p with coordinates 20 10 is the feasible solution of the constraint, point b with coordinates 25 0 is the feasible solution, point c is also the feasible solution. And any point outside the feasible region is an infeasible solution like the point d, point a the infeasible solutions. And also any point in the feasible region that gives the optimal value that is maximum or minimum value of the objective function is called an optimal solution. Now we have two very important theorems used in solving the linear programming problems. The first one says that if r be a feasible region for a linear programming problem and let z be equal to ax plus by the objective function then when z has an optimal value that is the maximum or minimum value where the variables x and y are subject to constraints described by inequalities, this optimal value must occur at corner point or you can say vertex of the feasible region. Next theorem says that if r be a feasible region for a linear programming problem and let z equal to ax plus by be the objective function then if r is bounded that is the feasible region is bounded and if r is bounded that is the feasible region is bounded then the objective function that is z has both a maximum and a minimum value on r and each of these occurs at a corner point or vertex of r that is the feasible region. Now we discuss the corner point method to solve the linear programming problem. In the first step what we do is we find the feasible region and determine the corner points. The corner points in this case is c, o, p and v and we have already found out the feasible region. That is the region o, b, p, c. Then in the next step we evaluate the objective function that is z equal to x plus y in this case at each corner point. So at the point c, 0, 20, z is equal to 20, at the point o, 0, 0, z is equal to 0, 0, 0, 0, 0, 0, 0, then at the point p, 20, 10, z is equal to 30 and at the point b, 25, 0, z is equal to 25. From these four values of z we see that the maximum value of z is 30 which is at x equal to 20 and y equal to 10. Now since the maximum value of z is 30 and we were supposed to find the maximum number of cakes, so we say that 30 cakes can be made from 5 kg of floor and 1 kg of fat. Now in this case the feasible region was bounded. Now we shall discuss the problem in which the feasible region is unbounded. Minimize z equal to x plus 2y subject to the constraints 2x plus y greater than equal to 3, x plus 2y greater than equal to 6, x greater than equal to 0 and y greater than equal to 0. Now we shall draw the graph and find the feasible region subject to these constraints. This shaded portion is the feasible region for this problem. Now we see that this feasible region is unbounded. These are the corner points of the feasible region that is the point a and point d, the value of z equal to x plus 2y at these corner points. So we have at a with coordinate 0, 3, z is equal to x plus 2y that is equal to 6 and at the point d with coordinate 6, 0, z is equal to x plus 2y and that is also equal to 6. Now as the feasible region here is unbounded, so 6 may or may not be the minimum value of z. So to decide this what we do is we graph the inequality x plus 2y less than 6. This green shaded portion is the region x plus 2y less than 6. Now let's see if the open half plane determined by x plus 2y less than 6 has any point common with the feasible region that is this yellow portion. We observe that there is no point common between these two regions. So 6 is the minimum value of z equal to x plus 2y. If in case we have a common point between the two regions then the objective function would have no minimum value. Now there is one more case when the feasible region is unbounded then m is the maximum value of the objective function if the open half plane determined by ax plus by greater than m has no point in common with the feasible region otherwise the objective function has no maximum value. We know that if two corner points of the feasible 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. So the minimum value of z occurs at all points on the line segment joining the points 0 common between the 3 and 6 comma 0. This completes the session hope you have understood what is the linear programming problem and how do we solve the linear programming problem by corner point method.