 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question phase. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7 per package on bolts. How many packages of each should be produced each day so as to maximize his profit if he operates his machines for the most 12 hours a day? Let us now start with the solution. First of all, let us assume that number of packages of nuts produced each day BX and number of packages of bolts produced each day by. So we can write let the number of packages of nuts and bolts produced each day BX and Y. Now we have given that machine A takes 1 hour to produce a package of nuts and machine A takes 3 hours to produce a package of nuts. And we also know that machines operate for at the most 12 hours a day. Now according to the question, X plus 3 Y is less than equal to 12. This is the time taken by machine A to produce X packages of nuts. And this is the time taken by machine A to produce Y packages of nuts. And we know machine operates for at the most 12 hours a day. So we get X plus 3 Y is less than equal to 12. Now we are also given that machine B takes 3 hours to produce a package of nuts and machine B takes 1 hour to produce a package of bolts. And this machine also operates for at the most 12 hours a day. Now our second constraint is 3X plus Y is less than equal to 12. This is the time taken by machine B to produce X packages of nuts. And this is the time taken by machine B to produce Y packages of nuts. So we get 3X plus Y is less than equal to 12. We know machine operates for at the most 12 hours a day. Now we also know that number of packages of nuts and packages of bolts produced per day is greater than equal to 0. So we can write X is greater than equal to 0 and Y is also greater than equal to 0. Now we are given that a profit of 57.50 is earned on one package of nuts and profit of 57 is earned on one package of nuts. Now our objective function Z is equal to 17.50 X plus 7.00 Y. This is the profit earned on X packages of nuts and this is the profit earned on Y packages of bolts. Now adding these two we get Z. Now our required real-life programming problem is maximize objective function Z is equal to 17.50 X plus 7 Y subject to constraints X plus 3 Y is less than equal to 12. 3X plus Y is less than equal to 12. X is greater than equal to 0 and Y is greater than equal to 0. Now to draw the graph and for finding the feasible region subject to given constraints first of all we will draw a line X plus 3 Y is equal to 12 corresponding to this inequality. Now these two points that is 0 4 and 12 0 lie on the line X plus 3 Y is equal to 12. So we will plug these two points on the graph and draw this line by joining these two points. Now this point on the graph represents the point 0 4 and this point represents 12 0. Joining these two points we get the line X plus 3 Y is equal to 12. Now clearly we can see this line divides the plane into two half planes. Now plane that contains the origin 0 0 satisfies the given inequality and we can say this plane represents X plus 3 Y is less than 12. So we will consider this plane. Now we will draw a line 3X plus Y is equal to 12 corresponding to this inequality. Now we know points 0 comma 12 and 4 comma 0 lie on the line 3X plus Y is equal to 12. Now plotting these two points 0 comma 12 and 4 comma 0 on the same graph and then joining them we get a line 3X plus Y is equal to 12. Now clearly we can see this line divides the plane into two half planes. Now this plane represents 3X plus Y less than 12. So we will consider this plane. As we can see plane containing the origin 0 0 represents 3X plus Y less than 12. Now we know X is greater than equal to 0 and Y is greater than equal to 0. This implies that the graph lies in the first quadrant only. Now we will consider all the half planes satisfying the given constraints and shading the common region we get this convex polygon. Now let us name this point as 0, coordinates of this point are 0 0. Now we will name this point as D, coordinates of point B are 0 comma 0. We will name this point as C. Clearly we can see these two lines intersect each other at point 3 comma 3. So coordinates of point C are 3 comma 3. Let us name this point as D, coordinates of D are 0 comma 4. Now according to the corner point method, maximum argument value of a linear objective function over a visible region determined by all the constraints occurs at some vertex of the polygon. Now clearly we can see these are the four vertices of the convex polygon. Now we can write the visible region ABCD determined by all the constraints is bounded. Now we will evaluate Z is equal to 17.5 X plus 7 Y at each corner point. Now first of all we will find the value of Z at corner point 0 comma 4. Z is equal to 28 at 0 comma 4 substituting 0 for X and Z for Y. In this function we get Z is equal to 28. Now we will find the value of Z at corner point 0 0. It is equal to 0 only. Clearly we can see substituting 0 for X and 0 for Y. In this expression we get Z is equal to 0 at 0 comma 0. Similarly the value of Z is equal to 70 at 4 comma 0 substituting 4 for X and 0 for Y. In this expression we get Z is equal to 70. Now we will consider corner point 3 comma 3. Now substituting 3 for X and 3 for Y in Z is equal to 17.5 X plus 7 Y we get Z is equal to 73.50 at point 3 comma 3. Naturally we can see this is the maximum value of Z which occurs at point 3 comma 3. So we can write hence the maximum value of Z is equal to 73.50 which occurs at point 3 comma 3. So we get maximum profit is equal to rupees 73.50. Number of packages of nodes produced daily is equal to 3 and number of packages of nodes produced daily is equal to 3. So this is our required answer. This completes the session. Have you understood the solution? Take care and have a nice day.