 Hello and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, a company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type A require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 3 hours for assembling. Profit is Rs.5 for type A and Rs.6 for type D souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit? Let us now start with the solution. First of all let us assume that number of type A souvenirs manufactured be X and number of type B souvenirs manufactured be Y. So we can write, let number of type A and type B souvenirs manufactured be X and Y respectively. Now we are given that time required for cutting type A souvenir is 5 minutes and time required for cutting type B souvenirs is 8 minutes. And there are 3 hours 20 minutes available for cutting. Now we get, according to the question, 5X plus 8Y is less than equal to 200. This is the time required to cut X souvenirs of type A and this is the time required to cut Y souvenirs of type B. Below there are 3 hours 20 minutes available for cutting and 3 hours 20 minutes is equal to 200 minutes. So sum of 5X minutes and 8Y minutes should be less than equal to 200. Now we are given time required for assembling type A souvenir is 10 minutes and time required for assembling type B souvenir is 8 minutes. Also we are given there are 4 hours available for assembling. Now our second constraint is 10X plus 8Y is less than equal to 240. Clearly we can see this is the time required for assembling X souvenirs of type A and this is the time required for assembling Y souvenirs of type B. And 4 hours is equal to 240 minutes. So 8X minutes plus 8Y minutes should be less than equal to 240. Below sum of these 2 times represent the total time of assembling and total time of assembling should be less than equal to 240. Now we can represent this inequality as 5X plus 4Y is less than equal to 120. Dividing both the sides of this inequality by 2 we get this equation. Now let us name this inequality as 1, this inequality as 2. Now we also know that number of souvenirs manufactured is greater than equal to 0. So we get X is greater than equal to 0 and Y is greater than equal to 0. We know X and Y represent the number of type A and type B souvenirs manufactured. Now we are also given that profit is rupees 5 each for type A and rupees 6 each for type B. Now our objective function becomes Z is equal to 5X plus 6Y. This represents the profit made on X souvenirs of type A and 6Y represents the profit made on Y souvenirs of type B. And sum of these 2 profits gives the profit function that is Z. Now the required LPP becomes maximize Z is equal to 5X plus 6Y, subject to constraints 5X plus 8Y is less than equal to 200, 5X plus 4Y is less than equal to 120, X is greater than equal to 0 and Y is greater than equal to 0. Here LPP denotes linear programming problem. Now to draw the graph and find the feasible region subject to given constraints. First of all we will draw a line 5X plus 8Y is equal to 200 corresponding to this inequality. Now we find that points 0 comma 25 and 40 comma 0 lie on the line 5X plus 8Y is equal to 200. Now this point represents 0 comma 25 and this point represents 40 comma 0. Joining these 2 points we take this line and this line represents 5X plus 8Y is equal to 200. Now this line divides the plane into 2 half planes. This plane satisfies 5X plus 8Y less than 200. So we will consider this plane. Now we will draw a line 5X plus 4Y is equal to 120 corresponding to this inequality on the same graph. Now we know points 0 comma 30 and 24 comma 0 lie on this line. Now plotting these 2 points on the graph and then joining these 2 points we get the line 5X plus 4Y is equal to 120. Clearly we can see this point represents 0 comma 30 and this point represents 24 comma 0. Joining these 2 points we get a line 5X plus 4Y is equal to 120. Now clearly we can see these 2 lines intersect at a point P here quadrants of P of 8 comma 20. Now again this line divides the plane into 2 half planes. Now this plane satisfies inequality 5X plus 4Y is less than 120. So we will consider this plane. Now we are also given that X is greater than equal to 0 and Y is greater than equal to 0. This implies the graph lies in the first quadrant only. Now clearly we can see this shaded portion in the graph is the feasible region satisfying all the given constraints. Now this shaded region is a convex polygon. This vertex of the polygon is O. Let us name this vertex as A. This vertex is P and this vertex let it be D. These are the corner points of the feasible region. Quadrants of O are 0,0. This is the origin. So it is coordinates are 0,0. Coordinates of point A are 24 comma 0. Coordinates of point P are 8 comma 20 and coordinates of point D are 0 comma 25. According to corner point method the maximum or minimum value of a linear function over feasible region determined by all the constraints occurs at some vertex of the polygon. Now we will find value of Z is equal to 5X plus 6Y at each arc of corner point. Now we can write feasible region OAPD is bounded. Therefore maximum value of Z will occur at any arc of corner points. Now let us evaluate value of objective function Z at each arc of corner point. Now first of all let us consider corner point 0 comma 25. Z is equal to 150 at 0 comma 25 substituting 0 for X and 25 for Y. In this equation we get value of Z is equal to 150. Now we will consider corner point 0,0. Value of Z at corner point 0,0 is equal to 0 only. Similarly we can find value of Z at corner point 24 comma 0. It is equal to 5 multiplied by 24 plus 6 multiplied by 0 which is further equal to 120. Now we will consider corner point 8 comma 20. Z is equal to 5 multiplied by 8 plus 6 multiplied by 20 which is further equal to 160. So we can write Z is equal to 160 at corner point 8 comma 20. Similarly we can see maximum value of Z is 160 that occurs at point 8 comma 20. Now we get company should manufacture 8 souvenirs of type A and 20 souvenirs of type B to make the maximum profit of rupees 160. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.