 Hello and welcome to the session. Let us discuss the following question. Question says, solve the following linear programming problem graphically minimize z is equal to x plus 2y subject to 2x plus y greater than equal to 3, x plus 2y greater than equal to 6, x greater than equal to 0, y greater than equal to 0. Let us now start with the solution. Now we have to minimize objective function. z is equal to x plus 2y subject to 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. Let us name these constraints as 1, 2, 3 and 4. Now to draw the graph and for finding the feasible region, subject to given constraints we shall first draw a line 2x plus y is equal to 3 corresponding to this inequality. Now points 0, 3 and 3 upon 2 comma 0 lie on the line 2x plus y is equal to 3. Now this point represents 0 comma 3 and this point represents 3 upon 2 comma 0. Joining these two points we take the line 2x plus y is equal to 3 on the graph. Now clearly we can see this line divides the plane into two half planes. Now this plane satisfies 2x plus y is greater than 3. So we will consider this plane or we can say plane which does not contains the origin 0, 0 that is this plane satisfies 2x plus y is greater than 3. Now we will draw a line x plus 2y is equal to 6 corresponding to this inequality. Now points 6, 0 and 0, 3 lie on the line x plus 2y is equal to 6. Now this point represents 0 comma 3 and this point represents 6 comma 0. Now joining these two points on the graph we take the line x plus 2y is equal to 6. Now again this line divides the plane into two half planes. We will consider the plane which does not contain 0, 0. We know this plane satisfies x plus 2y is greater than 6. We are also doing that x is greater than equal to 0 and y is greater than equal to 0. This implies graph lies in the first quadrant only. Now this shaded portion in this graph represents the feasible region. This region satisfies all the given constraints. Let us recall that the common region determined by all the constraints including the non-negative constraints x greater than equal to 0, y greater than equal to 0 of a linear programming problem is that feasible region. Now clearly we can see this feasible region is unbounded. Now we can write feasible region determined by the constraints 1 to 4 is unbounded. Now we will evaluate z at corner points. We know coordinates of corner points are 0 comma 3 and 6 comma 0. There are two corner points of this feasible region. First of all let us consider the corner point 0 comma 3. z is equal to 1 multiplying by 0 plus 3 multiplied by 2 which is further equal to 6. So we get z is equal to 6 at 0. 0 comma 3. We know z is equal to x plus 2y. Now substituting x and 3 for y in this equation we get the value of z at 0 comma 3. Similarly we can find the value of z at 6 comma 0. It is equal to 1 multiplied by 6 plus 2 multiplied by 0 which is further equal to 6 only. So we can write so it is equal to 6 at 0.6 comma 0. We know according to corner point method minimum value of z will occur at any of the two corner points. Now clearly we can say minimum value is same at both the corner points. We know feasible region is unbounded. So 6 is the minimum value of objective function z is open half plane z is less than 6 does not have any point common with the feasible region. Now clearly we can see this down shaded region represents x plus 2y less than 6 or we can say this region represents z less than 6. Clearly we can see these two planes have no common point or we can say this region has no point common with the feasible region. Now we can write z less than 6 does not have any point common with the feasible region. Hence the minimum value of z is equal to 6. Now we also know that if two corner points of the feasible region produce the same maximum or minimum value then any point line on the line segment joining these points is also an optimal solution of the same type. Now clearly we can see this line joins these two points. So minimum value of z that is 6 is accurate all the points on this line. So we can write minimum value of z is equal to 6 accurate all the points on the line segment joining the points 6,0 and 0,3. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.