 Hello and welcome to the session. Let us discuss the following question. Question says, a cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding or cutting machine and a sprayer. It takes 2 hours on grinding or cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding or cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding or cutting machine for at the most 12 hours. The profit from the sale of the lamp is Rs 5 and that from the shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit? Let us now start with the solution. Now we shall first formulate the linear programming problem according to the given conditions in the question and then solve it graphically. First of all, let us assume that let number of pedestal lamp produced be X. So we can write let number of pedestal lamps produced be X and also let the number of shades produced be Y. Now we are given that it takes 2 hours on grinding or cutting machine to produce 1 lamp and it takes 1 hour on grinding and cutting machine to produce 1 shade. We are also given that grinding or cutting machine is available for at the most 12 hours. Now we can write according to the given question 2X plus Y is less than equal to 12. This is the time taken by grinding or cutting machine to produce X pedestal lamps and this is the time taken by grinding or cutting machine to produce Y number of shades and we know this total time is less than equal to 12 as machine is available for at the most 12 hours. Now this is our required first constraint. Now we are given that it takes 3 hours on the sprayer to manufacture a pedestal lamp and 2 hours on the sprayer to manufacture a shade and we also know that sprayer is available for at the most 20 hours. Now our required second constraint is 3X plus 2Y is less than equal to 20. Clearly we can see this is the time taken by sprayer to produce X pedestal lamps and this is the time taken by sprayer to produce Y shades. Now total time is less than equal to 20 as sprayer is available for at the most 20 hours. Now we know the number of lamps and shades produced is greater than equal to 0. So we can write X is greater than equal to 0 this is our third constraint and Y is greater than equal to 0 this is our fourth constraint. Now we are also given in the question that the profit from the sale of the lamp is rupees 5 and that from the shade is rupees 3. Now profit function or we can say objective function Z is equal to 5X plus 3Y this is the total profit made on X number of lamps and this is the total profit made on Y number of shades adding these two profits we get a profit function that is Z. Now linear programming problem becomes maximize Z is equal to 5X plus 3Y subject to the constraints 2X plus Y is less than equal to 12, 3X plus 2Y is less than equal to 20, X is greater than equal to 0 and Y is greater than equal to 0. Now for drawing the graph and for finding the feasible region subject to given constraints we shall first draw a line 2X plus Y is equal to 12 corresponding to this inequality. Now clearly we can see points 0 comma 12 and 6 comma 0 lie on the line 2X plus Y is equal to 12. Now graph of the line can be drawn by plotting these two points on the graph and then joining them now clearly we can see in the graph this is the point 0 comma 12 and this is the point 6 comma 0 joining these two points we get the line 2X plus Y is equal to 12. Now clearly we can see this line divides the plane into two half planes now we will consider the half plane that satisfies 2X plus Y is less than 12 that is the half plane containing 0 0. Now we will draw a line 3X plus 2Y is equal to 20 corresponding to this inequality now clearly we can see points 0 comma 10 and 6 comma 1 lie on the line 3X plus 2Y is equal to 20. Now we will plot these two points on the same graph and then we will join them to get this line now clearly we can see this point represents 0 comma 10 and this point represents 6 comma 1 joining these two points we get the line 3X plus 2Y is equal to 20 now again this line divides the plane into two half planes we will consider the plane that satisfies the inequality 3X plus 2Y is less than 20 or we can say we will consider the plane that contains origin 0 0 also clearly we can see lines 2X plus Y is equal to 12 and the line 3X plus 2Y is equal to 20 intersect each other at point 4 comma 4 now we have given X is greater than equal to 0 and Y is greater than equal to 0 now this implies that the graph lies in the first quadrant now this shaded region in the graph represents the feasible region satisfying all the given constraints now clearly we can see feasible region is represented by this convex polygon let us name this point as C this point as P and this point as B so we get convex polygon C O B P represents the feasible region corner points of this convex polygon are O having coordinates 0 0 B having coordinates 0 10 P having coordinates 4 comma 4 and C having coordinates 6 comma 0 now we can write a shaded portion in the graph which is a convex polygon C O B P represents feasible region now the corner points of the convex polygon C O B P are B 0 comma 10 O 0 comma 0 C 6 comma 0 and P 4 comma 4 now according to corner point method maximum or minimum value of a linear objective function over a convex polygon occurs at some vertex of the polygon now we can write according to corner point method maximum value of Z will occur at any of the above points now we know Z is equal to 5 X plus 3 Y first of all let us find the value of Z at point 0 comma 10 Z is equal to 30 at 0 comma 10 substituting 0 for X and 10 for Y in this expression we get Z is equal to 30 at point 0 comma 10 similarly Z is equal to 0 at point 0 comma 0 now we will find the value of Z at point 6 comma 0 substituting 6 for X and 0 for Y in this expression we get Z is equal to 30 now we will find the value of Z at point 4 comma 4 so substituting 4 for X and 4 for Y in this expression we get Z is equal to 32 now clearly we can see maximum value of Z is 32 and it occurs at point 4 comma 4 so we can write hence the maximum value of Z is equal to 32 which occurs at point 4 comma 4 that is value of X is equal to 4 and value of Y is also equal to 4 now we know X is the number of pedestrian lamps so we get number of pedestrian lamps is equal to 4 and Y is the number of shades so we can write number of wooden shades is equal to 4 and maximum profit is equal to rupees 32 so this is our required answer this completes the session hope you understood the solution take care and have a nice day