 Hello and welcome to the session. In this session we discussed the final question which says a small form manufactures one rings and chains. The total number of rings and chains manufactured per day is at most 24. It takes 1 hour to make a ring and 30 minutes to make a chain. The maximum number of hours available per day is 16. If the profit on a ring is Rs 300 and that on a chain is Rs 190, find the number of rings and chains that should be manufactured per day so as to earn the maximum profit, make it an LPP and solve it graphically. Let us now proceed with the solution. Let us write the given data in the tabular form. Like in the question it is given that a small form manufactures gold rings and chains. So the two items would be rings and chains. It is given that it takes 1 hour to make a ring and 30 minutes to make a chain. So time taken to make a ring is written as 1 and to make chain 30 minutes are taken which means half an hour. We are also given that the profit on a ring is Rs 300. So we write here 300 and profit on the chain is Rs 190. We write here 190. As we are supposed to find the number of rings and chains that should be manufactured per day. So we can take the number of rings manufactured per day would be say y. So we suppose that the number of rings the x units the number of chains may be y units. Per day and y chains per day to maximize the profit is given in the table that a profit on a ring is Rs 300. Would be equal to Rs 300x when one chain is Rs 190. So the total profit is equal to Rs 190 high. Also given that the total number of rings and chains manufactured per day is at most 24. So this means we have x equal to 24 where x is the number of rings manufactured per day and y is the number of chains manufactured per day. So x plus y plus n equal to 24. We have obtained one constraint for the linear programming problem. As in the fashion it's given that it takes 1 hour to make a ring and half an hour to make a chain. So it's equal to half into y that is equal to 1 upon 2 into y. Also in the question we are given that the maximum number of hours available per day y is less than equal to 16. Thus we have obtained another constraint and this could also be written as it is less than equal to 32. Rings cannot be negative. So x is also greater than equal to 0 and y is also greater than equal to 0. Now the mathematical formulation ring problem is given as the total profit that is say z equal to Rs 30 by z equal to y given as less than equal to 24 and the next constraint is 2x plus y less than equal to 32. With this equation 2 we have the non-negative constraints as equal to 0, greater than equal to 0 like this equation 3. So we have made a linear programming problem. Now we have to solve this problem graphically. The first step of solving this linear programming problem graphically is in the physical region determined by these constraints. So for this first we need to graph this inequality for which we will graph the equation x plus y equal to 24. Now putting 2 to 0, y equal to 0 in inequality run we get 0 plus 0 is less than equal to 24 that is 0 is less than equal to 24 which is true the origin which coordinates 0 to 0 like in the region x plus y less than equal to 24. So this region this shaded portion below the line x plus y equal to 24 represents the inequality x plus y less than equal to 24. Now next we will graph the inequality 2x plus 32 of the equation 2x plus y equal to 32 which represents the equation. Now we will check if the origin lies in the region or this putting 0, y equal to 0 in equation 2 into 0 plus 0 is less than equal to 32 that is 0 is less than equal to 32 which is therefore we say that the origin 0 0 lies in the region given by the inequality 2x plus y less than equal to 32. This shaded portion in red represents the region quality 2x plus y less than equal to 32. We have graphed the inequalities of the constraints x plus y less than equal to 24 and 2x plus y less than equal to 32. Let's now consider the constraints x greater than equal to 0 and y greater than equal to 0. Now the non-negative constraint x greater than equal to 0 is the y axis in the region right hand side and y greater than equal to 0 is the x axis and the region to the right of the y axis is x greater than equal to 0 and the x axis and the region above the x axis is the region y greater than equal to 0. Now let's determine the common region determined by all these constraints. The section of the line joining the points c and d that is the line 2x plus y equal to 32 with the x axis that is this point v point e region e b b o e the feasible region. The last time the corner points of the feasible region which are the points e, d, b and o the corner points the region as passes the point e with coordinates 16, 0 then we have the point d with coordinates point b with coordinates 0, 24 is 0, 0. The value of the objective function given by z equal to 300x plus 190y which is the total profit corner point e with coordinates 16, 0. So for this the value of the objective function z would be equal to 300x16 to 0 and so this would be equal to 800 which is the point d with coordinates 8, 16. The value of the objective function z is equal to 300x8 plus 190x16 be equal to 5,440. The point is the point b with coordinates 0, 24 so z would be equal to 300 into 0 plus 190 into 24 and so this would be equal to 4560 the point o with coordinates 00. The value of the objective function z would be equal to 300 into 0 plus 190 into 0 and so this would be equal to 0. Of these values of the objective function obtained at the corner points of the feasible region we find that minimum value and this is the minimum value. So we take the maximum value as capital M equal to 5,440 and the minimum value small n equal to 0. So we have the maximum value to profit function or you can say the objective function as in 440 at the corner point d with coordinates 8, 16. We are supposed to maximize the profit so we will consider only the maximum value of the objective function z. As we can now say the maximum profit would be rupees 440 at manufacturers that assumed x to be the number of rows manufactured by the firm per day and y to be the number of chains manufactured by the firm per day. As we say that if the firm manufactured then the maximum profit would be rupees 5,440. So this is our final answer. This completes the question. Hope you understood the solution of this question.