 Hello, I am welcome to the session. I am Deepika here. Let's discuss the question which says A factory makes tennis racquets and cricket bats. A tennis racquet takes 1.5 hours of machine time and 3 hours of craftsman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman's time. In RT, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman's time. Part 1. What number of racquets and bats must be made if the factory is to work at full capacity? Part 2. If the profit on a racquet and on a bat is Rs. 20 and Rs. 10 respectively, find the maximum profit of the factory when it works at full capacity. So, let's start the solution. Now, in this question we will first formulate the linear programming problem according to the given conditions and then we will solve it graphically. Let X is the number of tennis racquets and Y is the number of cricket bats. It is given that a tennis racquet takes 1.5 hours of machine time and a cricket bat takes 3 hours of machine time and the total machine time available is not more than 42 hours. So, we have 1.5X plus 3Y less than equal to 42. So, this is a constraint related to machine time or this can be written as 3X plus 6Y less than equal to 84 or X plus 2Y less than equal to 28. So, this is our first constraint. Again, we are given a tennis racquet takes 3 hours of craftmen's time while a cricket bat takes 1 hour of craftmen's time. Also, the total craftmen's time available is not more than 24 hours. So, again we have 3X plus Y less than equal to 24. So, this is our second constraint which is related to craftmen's time. Also, the number of racquets and bats made is greater than equal to 0. So, we have X greater than equal to 0 and Y greater than equal to 0. So, these are our non-negative constraints. Again, the profit on a racquet and on a bat is rupees 20 and rupees 10 respectively. Now, the total profit in rupees is equal to 20X plus 10Y. Let Z is equal to 20X plus 10Y. Hence, the mathematical formulation the problem is maximize Z is equal to 20X plus 10Y subject to the constraints 2Y less than equal to 28. This is our machine time constraint and 3X plus Y less than equal to 24. This is our craftmen's time constraint and X greater than equal to 0 and Y greater than equal to 0. These are the non-negative constraints. So, we have to maximize Z is equal to 20X plus 10Y. Let us take this as number 1 subject to these constraints. Let us take these constraints as 2, 3 and 4. Now, we will first draw the graph and find the feasible region subject to these given constraints. Now, the equation of the line corresponding to the inequality X plus 2Y less than equal to 28 is X plus 2Y is equal to 28. So, we will first draw the line representing the equation X plus 2Y is equal to 28. Now, we observe that the points 0, 14 and 28, 0 lie on the line X plus 2Y is equal to 28. Therefore, the graph of the line can be drawn by plotting points 0, 14 and 28, 0 and then joining them. Now, let us take A as a point, 0, 14 and B as a point, 28, 0. So, AB is the line representing the equation X plus 2Y is equal to 28. Now, the line AB divides the plane into two half planes. Now, clearly the origin does not lie on this line. It lies on the half plane of 2. So, the closed half plane containing the origin is the graph of 2. Now, again the equation of the line corresponding to the inequality 3X plus Y less than equal to 24 is 3X plus Y is equal to 24. So, now we will draw the line representing the equation 3X plus Y is equal to 24 on the same graph. Clearly the points 0, 24 and 80 lie on the line 3X plus Y is equal to 24. So, we will plot these points on the graph and then join them. Let us take C as a point, 0, 24 and D as a point, 80. So, CD represents the equation 3X plus Y is equal to 24. Now, again the line CD divides the plane into two half planes. So, the half plane containing the origin satisfies 3. So, the closed half plane containing the origin is the graph of 3. Again, X greater than equal to 0 and Y greater than equal to 0 implies that the graph lies in the first coordinate only. Now, here the line AB and CD intersect at a point. Let us take this point as the point P. So, the coordinates of P are 4, 12 that is P is the point whose X coordinate is 4 and 5 coordinate is 12. Now, the shaded region in the graph is the feasible region satisfying all the given constraints. Clearly, the shaded region is bounded. So, we will apply the corner point method to determine the maximum value of set. Now, the coordinates of corner points A, O, D and P 0, 14, 0, 0, 8, 0 and 4, 12 respectively. Now, according to the corner point method, maximum value of Z will occur at any of these points. So, let us evaluate the objective function Z at each corner point. So, at the point A with coordinates 0, 14, Z is equal to 20 into 0 plus 10 into 14 and this is equal to 0 plus 140 which is equal to 140. Now, at origin Z is equal to 20 into 0 plus 10 into 0 which is equal to 0. Again, at the point D with coordinates 8, 0, Z is equal to 20 into 8 plus 10 into 0 which is equal to 0. Now, at the point P with coordinates 4, 12, Z is equal to 20 into 4 plus 10 into 12 and this is equal to 80 plus 120 which is equal to 200. Hence, the maximum value of Z is equal to 200 which occurs when X is equal to 4 and Y is equal to 12. Hence, the factory must make 4 tennis records and 12 cricket bats to realize maximum profit and maximum profit then will be rupees 200. Hence, the answer for part 1 is 4 tennis records, 12 cricket bats and for part 2 is maximum profit is equal to rupees 200. So, this completes our session. I hope the solution is clear to you. Bye and have a nice day.