 Good morning friends. I am Poorva and today we will work out the following question. A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time required for each toy on the machine is given below. Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit. Let us begin with the solution now. Now we shall first formulate the linear programming problem according to the given conditions and then solve it. Now we have to show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit. Now let number of type A toys manufactured be equal to x and number of type B toys manufactured be equal to y and we have to show that x is equal to 15 and y is equal to 30. Now we are given that three machines are needed for manufacturing the toys and the time required for each toy is given in the following table. We are also given the maximum time available for each machine to be 360 minutes. From the table we can clearly see that machine 1 takes 12 minutes to manufacture a toy of type A and 6 minutes to manufacture a toy of type B. Machine 2 takes 18 minutes to manufacture a toy of type A and 0 minutes to manufacture a toy of type B and machine 3 takes 6 minutes to manufacture a toy of type A and 9 minutes to manufacture a toy of type B. And we are also given that each machine is available for a maximum of 360 minutes. Now we get the following inequalities from the given data. Now number of type A toys manufactured is equal to x and machine 1 takes 12 minutes to manufacture one toy of type A. So to manufacture x toys it will take 12 into x minutes and number of type B toys manufactured is equal to y and machine 1 takes 6 minutes to manufacture one toy. So to manufacture y toys it will take 6 into y minutes. So the total time is 12x plus 6y and this should be less than equal to 360 because machine 1 is available for a maximum of 360 minutes. And this implies 2x plus y is less than equal to 60. This is our constrained one. Now again number of type A toys manufactured is x and machine 2 takes 18 minutes to manufacture one toy. So to manufacture x toys it will take 18 into x minutes and number of type B toys manufactured is y and machine 2 takes 0 minutes to manufacture one toy. So to manufacture y toys it will take 0 into y minutes. So the total time is 18x plus 0y and this is less than equal to 360 because machine 2 is available for a maximum of 360 minutes. Now this implies x is less than equal to 20. This is our constrained 2. Now number of type A toys manufactured is x and machine 3 takes 6 minutes to manufacture one toy. So to manufacture x toys it will take 6 into x minutes and number of type B toys manufactured is y and machine 3 takes 9 minutes to manufacture one toy. So to manufacture y toys it will take 9 into y minutes. So the total time is 6x plus 9y and this is again less than equal to 360 because machine 3 is available for a maximum of 360 minutes. Now this implies 2x plus 3y is less than equal to 120. This is our constrained 3. Now according to the given problem we have to maximize the profit and we are given that the profit on each toy of type A is rupees 7.50 and that on each toy of type B is rupees 5. So we get the objective function as z is equal to 7.5x plus 5y. Now number of type A toys manufactured is x and profit on one type A toy is rupees 7.50. So profit on x toys will be 7.5 into x and number of type B toys manufactured is y and profit on one type B toy is rupees 5. So on y toys the profit will be 5 into y. So the total profit is 7.5x plus 5y and so we get the objective function as z is equal to 7.5x plus 5y. Also we have x greater than equal to 0 and y greater than equal to 0 because the number of type A and type B toys manufactured is greater than equal to 0. Hence the linear programming problem becomes maximize z is equal to 7.5x plus 5y subject to the constraints 2x plus y less than equal to 60. x less than equal to 20, 2x plus 3y less than equal to 120, x greater than equal to 0 and y greater than equal to 0. Now the first inequality is 2x plus y less than equal to 60 and the line corresponding to this inequality is 2x plus y equal to 60 and the points 0, 60 and 30, 0 lie on this line. So for drawing the graph and finding the feasible region subject to the given constraints we shall first draw the line representing the equation 2x plus y equal to 60 corresponding to the inequality 2x plus y less than equal to 60 on the graph by plotting the points 0, 60 and 30, 0 and then joining them. So we have plotted the points 0, 60 and 30, 0 on the graph join them and we have named the line as AB. This line AB divides the plane into two half planes we shall take the half plane that satisfies 2x plus y less than 60 that is the portion of the graph below this line along with the line AB in the feasible region. Now the second inequality is x less than equal to 20 and the equation corresponding to the inequality x less than equal to 20 is x is equal to 20, x is equal to 20 represents a line parallel to y axis on the graph which passes through 20, 0. So this is the line x is equal to 20 which is parallel to y axis and which passes through 20, 0. We have named this line as CD. This line CD divides the plane into two half planes we shall take the half plane satisfying x less than 20 that is the portion of the graph to the left inside of this line along with the line CD in the feasible region. Now the third inequality is 2x plus 3y less than equal to 120 and line corresponding to this inequality is 2x plus 3y equal to 120 and the points 0, 40 and 60, 0 lie on this line. So now we shall represent 2x plus 3y less than equal to 120 graphically by drawing the line 2x plus 3y equal to 120 by plotting the points 0, 40 and 60, 0 on the same graph and then joining them. So we have joined the points 0, 40 and 60, 0 and named the line as EF. This line EF divides the plane into two half planes we shall take the half plane satisfying 2x plus 3y less than 120 that is the portion of the graph below this line EF along with the line EF in the feasible region. Also x greater than equal to 0 and y greater than equal to 0 implies that the graph lies in the first quadrant only. The lines A, B, CD and EF intersect each other at points L, M and N. Thus the shaded portion in the graph is the feasible region satisfying all the given constraints. Here the feasible region is the convex polygon EODML with coordinates of vertices as 0, 40, 0, 0, 20, 0, 20, 20 and 15, 30. So according to the corner point method which states that the maximum or minimum value of a linear objective function over a convex polygon occurs at some vertex of the polygon. The maximum value of Z will occur at any of the above points. Therefore we will calculate the value of Z is equal to 7.5x plus 5y at these points. So Z is equal to 7.5 into 0 plus 5 into 40 at 0, 40 and this comes out to be equal to 200. Z is equal to 7.5 into 0 plus 5 into 0 at 0, 0 and this comes out to be equal to 0. Z is equal to 7.5 into 20 plus 5 into 0 at 20, 0 and this comes out to be equal to 150. Z is equal to 7.5 into 20 plus 5 into 20 at 2020 and this comes out to be equal to 250. And finally Z is equal to 7.5 into 15 plus 5 into 30 at 1530 and this comes out to be equal to 262.50. Hence the maximum value of Z out of these five values is equal to 262.50 which occurs at 1530 or we can say profit is maximum when x is equal to 15 and y is equal to 30. Thus 15 toys of type A and 30 toys of type B should be manufactured to get the maximum profit. This is our answer. Hope you have understood the solution. Bye and take care.