 Hello and welcome to the session, let's discuss the following question. It says solve the following linear programming problem graphically minimize z equal to x minus 5y plus 20 subject to the constraints x minus y greater than equal to 0 minus x plus 2y greater than equal to 2, x greater than equal to 3, y less than equal to 4 and x and y are greater than equal to 0. So, let's now move on to the solution. We have to minimize z equal to x minus 5y plus 20 subject to the constraints x minus y greater than equal to 0 minus x plus 2y greater than equal to 2, x greater than equal to 3, y less than equal to 4 and x and y are greater than equal to 0. Now the first step towards solving this linear programming problem is to draw the lines corresponding to the constraints. Now we draw the line x minus y is equal to 0 that is x is equal to y. Well first of all draw this line x is equal to y. So, x is equal to y is the line passing through the origin. Now we have to draw the line minus x plus 2y is equal to 2 for that we make a table which gives us the points through which the line passes. If x is 0 then y is 2 by 2 that is 1 and if y is 0 x is minus 2. So, now we draw the line minus x plus 2y is equal to 2 when x is 0 y is 1 that is this point when y is 0 x is minus 2 that is this point. Now we join these two points to get the line minus x plus 2y is equal to 2. Now we have to draw the line corresponding to the inequality x greater than equal to 3 and the line corresponding to the inequality x greater than equal to 3 is x is equal to 3. So, now here we have x is equal to 3. Now we draw the line x equal to 3 this is the line x is equal to 3. Now we have to draw the line corresponding to the constraint y less than equal to 4 and the line corresponding to the constraint y less than equal to 4 is y is equal to 4. So, here we have y is equal to 4. So, now we draw the line corresponding y is equal to 4 this is the line y is equal to 4. Now we will shade the region satisfying all the constraints. Now the first constraint is x is greater than equal to 0. So, this implies x is greater than equal to y. So, we have to shade the region in which x is greater than equal to y. So, this is the line x is equal to y and we have to shade the region which gives us all the values of x which are greater than equal to y. Now here we see that in this region in fact this whole region that is the region below the line x is equal to y is the region which satisfies that which satisfies the condition that x is greater than equal to y. Now we have to shade the region corresponding to the constraint minus x plus 2y greater than equal to 2. Now here to check that which region satisfies this inequality will take any values of x and y which does not lie on the line minus x plus 2y is equal to 2 and we will see that whether that point satisfies this inequality or not. If that point satisfies this inequality will shade the region which contains that point and if that point doesn't satisfies this inequality will shade the region which doesn't contain the point. So, here we take x and y as 0 0 so 0 plus 0 is greater than equal to 2 which is not true. So, we have to shade the region which does not contain the point 0 0 for the inequality minus x plus 2y greater than equal to 2. So, we have to shade this region because this region does not contain the point 0 0. Now we have to shade the region for the inequality x greater than equal to 3. So, this is the region which satisfies the inequality x greater than equal to 3 and also we have to shade the region for the inequality y less than equal to 4. So, this is the region the region below the line y is equal to 4 is the region satisfying the inequality y less than equal to 4 and And we will see that this is the common region satisfying all the constraints. So let's name this region A, B, C, D. So A, B, C, D is the feasible region. Now we need to find the vertices of the feasible region. Now here, y is equal to 4 and this is the line x is equal to y. So since this line is y is equal to 4 and x is equal to y, so this point is 4, 4 and this point B is the point of intersection of the line, x is equal to y and x is equal to 3. So B has coordinates 3, 3 since here x is equal to 3 and this line is x is equal to y. So coordinates of the point B are 3, 3 and here the x coordinate is 3 since this is the line x is equal to 3 and this is the point of intersection of the line x is equal to 3 and minus x plus 2y is equal to 2. So to find the value of y, we will put x is equal to 3 here. So if we put x is equal to 3 here, we have minus 3 plus 2y is equal to 2. So we have 2y is equal to 3 plus 2 that is 5. So this implies y is equal to 5 by 2 and here y is equal to 4 and D is the point of intersection of the line minus x plus 2y is equal to 2 and the line y is equal to 4. So here we will put y is equal to 4. So we have minus x plus 2 into 4 is equal to 2. So this implies x is equal to 8 minus 2. So this implies x is equal to 6. So the coordinates of the point D are 6, 4. x coordinate is 6, y coordinate is 4. So A is 4, 4, B is 3, 3. C is 3, 5 by 2. D is 6, 4. So these are the vertices of the feasible region. Now when x is 4, y is 4 that is at A, z is equal to x minus 5y plus 20 which is equal to 4 minus 5 into 4 plus 20 which is equal to 5 minus 20 plus 20 which is equal to 4. Now at B this 3, 3, z is equal to 3 minus 5 into 3 plus 20 which is equal to 3 minus 15 plus 20 which is equal to 8 and at C, x is 3, y is 5 by 2, z is equal to 3 minus 5 into 5 by 2 plus 20 which is equal to 6 minus 25 plus 20 plus 40 by 2 which is equal to 21 by 2 which is equal to 10.5. Now at D, x is 6, y is 4. So z is equal to 6 minus 5 into 4 plus 20 which is equal to 6 minus 20 plus 20 which is equal to 6. So we see that at A, z is minimum that is when x is 4, y is 4, z is minimum it has the least value 4 and we have to minimize z. Therefore z is minimum when x is equal to 4 and y is equal to 4. So x is equal to 4 and y is equal to 4 is the required answer that is to minimize z. So this completes the question and the session. Bye for now. Take care. Have a good day.