 Good morning friends. I am Purva and today we will work out the following question. A factory manufactures two types of screws A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes four minutes on the automatic and six minutes on hand operated machines to manufacture a package of screws A. While it takes six minutes on automatic and three minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most four hours on any day. The manufacturer can sell a package of screws A at a profit of Rs.7 and screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit, determine the 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 in the question we are given that assuming that the manufacturer can sell all screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximize his profit. So let number of packages of screws A produced B equal to X. Number of packages of screws B produced B equal to Y. Now in the question we are given that it takes four minutes on the automatic and six minutes on hand operated machines to manufacture a package of screws A. While it takes six minutes on automatic and three minutes on the hand operated machines to manufacture a package of screws B. And we are also given that each machine is available for at the most four hours on any day and the manufacturer can sell a package of screws A at a profit of Rs.7 and screws B at a profit of Rs.10. So we make the following table from the given data. Now from this table we can clearly see that the time required on automatic machine to produce a package of screws A is four minutes and a package of screws B is six minutes. While the time required on hand operated machine to produce a package of screws A is six minutes and a package of screws B is three minutes. We can also see that the maximum time for which automatic and hand operated machines are available is 240 minutes. We can also see that on selling a package of screws A the manufacturer gets a profit of Rs.7 and on selling a package of screws B the manufacturer gets a profit of Rs.10. So from the given information we can find the inequalities. Now number of packages of screws A produced is equal to X and time taken on automatic machine to produce a package of screws A is four minutes. So to produce X packages the time taken is four into X minutes and number of packages of screws B produced is equal to Y. And time taken on automatic machine to produce one package of screws B is six minutes so to produce Y packages the time taken is six into Y minutes. So the total time taken is four X plus six Y and this is less than equal to 240 because we are given that each machine is available for a maximum of 240 minutes. This implies two X plus three Y is less than equal to 120 this is our constraint one. Now again time taken on hand operated machine to produce a package of screws A is six minutes so to produce X packages the time taken is six into X minutes. And the time taken on hand operated machine to produce a package of screws B is three minutes so to produce Y packages the time taken will be three into Y minutes. So the total time taken is six X plus three Y and this is less than equal to 240 because we are given that each machine is available for a maximum of 240 minutes. Now this implies two X plus Y is less than equal to 80 this is our constraint two. Also we have X greater than equal to zero and Y greater than equal to zero as the number of packages of screws A and B produced is greater than equal to zero. Now according to the given problem we have to maximize the profit so we get the objective function as Z is equal to seven X plus ten Y. Now the profit on one package of screws A is rupees seven so on X packages the profit will be seven into X and the profit on one package of screws B is rupees ten so on Y packages the profit will be ten into Y. So the total profit is seven X plus ten Y and so we get the objective function as Z is equal to seven X plus ten Y. Hence the linear programming problem becomes maximize Z is equal to seven X plus ten Y subject to the constraints two X plus three Y less than equal to 120 two X plus Y less than equal to 80 X greater than equal to zero and Y greater than equal to zero. Now the first inequality is two X plus three Y less than equal to 120 and the line corresponding to this inequality is two X plus three Y equal to 120 and the points zero forty and sixty zero 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 two X plus three Y equal to 120 corresponding to the inequality two X plus three Y less than equal to 120 on the graph by plotting the points zero forty and sixty zero and then joining them. So we have plotted the points zero forty and sixty zero on the graph joined them and we have named the line as AB. Now this line AB divides the plane into two half planes we shall take the half plane that satisfies two X plus three Y less than 120 that is the portion of the graph below this line AB along with the line AB in the feasible region for this inequality. Now the second inequality is two X plus Y less than equal to 80 and the line corresponding to this inequality is two X plus Y equal to 80 and the points zero eighty and forty zero lie on this line. So now we shall draw the line representing the equation two X plus Y equal to 80 on the same graph by plotting the points zero eighty and forty zero which satisfy the equation and then joining them. So we have plotted the points zero eighty and forty zero on the same graph joined them and we have named the line as CD. Now this line CD divides the plane into two half planes we shall take the half plane that satisfies two X plus Y less than 80 that is the portion of the graph to the left hand side of this line CD along with the line CD in the feasible region for this inequality. Also we have X greater than equal to zero and Y greater than equal to zero which implies that the graph lies in the first quadrant only. The lines AB and CD intersect each other at the point P. Thus the shaded portion in the graph is the feasible region satisfying all the given constraints. Here the feasible region is a convex polygon A O D P with vertices zero forty zero zero forty zero and thirty twenty. 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. Maximum value of Z will occur at any of the above points therefore we will calculate the value of Z is equal to 7 X plus 10 Y at these points. Now Z is equal to 7 X zero plus 10 X forty at zero forty and this comes out to be equal to 400 Z is equal to 7 X zero plus 10 X zero at zero zero and this comes out to be equal to zero. Z is equal to 7 X 40 plus 10 X zero at 40 zero and this comes out to be equal to 280 and Z is equal to 7 X 30 plus 10 X 20 at 30 20 and this comes out to be equal to 410. Hence the maximum value of Z out of these four values is 410 which occurs at 30 20. Now this implies X is equal to 30 and Y is equal to 20. So we get that maximum profit is equal to rupees 410 and number of packages of screws A is equal to 30 and number of packages of screws B is equal to 20. So we write our answer as 30 packages of screws A and 20 packages of screws B and maximum profit is equal to rupees 410. This is our answer. Hope you have understood the solution. Bye and take care.