 Hi and how are you all today? The question says solve the following linear programming problem graphically. Now this is the second question that we need to proceed on, says minimize z is equal to 3x plus 5y subject to the constraint, x plus 3y is greater than equal to 3, x plus y is greater than equal to 2 and the value of x and y is obviously greater than equal to 0. Now here we will be using the fundamental extreme point theorem to proceed on with our solution. As we are well versed with this theorem, let us straightaway start with our solution. Now first of all we need to write down the equation corresponding to x plus 3y is greater than equal to 3 and that is x plus 3y is equal to 3. Now the points line on it are given by, let us find out two points that will satisfy this line. When we substitute x as 0 we have the value of y as 1 whereas when we substitute the value of y as 0 we have the value of x as 3. Then the same procedure will be for equation corresponding to x plus y is greater than equal to 2 and that is x plus y is equal to 2. Let us again find out two points, value of x is 0 then the value of y is 2 or if we substitute value of y as 0 we have the value of x as 2. Now we shall draw these two lines on the graph. Also we know that since x is greater than equal to 0 and the value of y is also greater than equal to 0 this implies that the graph lies in the first quadrant. Now let us draw the graph for this. Here when the value of x is 0 then the y's value is 1 so this is a point that is 0, 1. When the value of x is 3 then the value of y is 0 so on joining these two points we will be representing the line that is x plus 3y is equal to 3. Now we shall draw the line representing the equation x plus y is equal to 2 by the same method. Here when x is 0 y is 2 then y is 0 x is that is join these two points also and this line is representing x plus y is equal to 2. Here this shaded portion in the graph is the feasible region satisfying all the constraints. It is an unbound from top and the corner points of the lower region are 0, 2. This point is found out to be 3 by 2 comma 1 by then here it is 3 is 0. According to the fundamental extreme point term the minimum value of z will occur at any one of these three points. Now let us calculate the value of z at these points. The shaded region is the feasible region corresponding to x plus 3y is greater than equal to 3, x plus y is greater than equal to 2 and x comma y is greater than equal to 0 and the corner points are 0 comma 2 3 by 2 comma 1 by 2 3 comma 0. Finally we have to minimize the value of z that is 3x plus 5y. Now at 0 comma 2 the value of z is equal to 10 3 by 2 comma 1 by 2 the value of z is equal to 7 at 3 comma 0 the value of z is equal to 9. So we can write that therefore minimum value of z that is our objective function is 7 when the value of a 3 by 2 and the value of y is equal to 1 by 2. So this is our final answer to this session. Hope you enjoyed drawing the graph and understood the concept well. Have a good day.