 Hello and welcome to the session, I am Deepika here. Let's discuss a question which says, A merchant plans to sell two types of personal computers, a desktop model and a portable module that will cost rupees 25,000 and rupees 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units, determining the number of units of each type of computers which the merchant should stop to get maximum profit if he does not want to invest more than rupees 70 lakhs and if his profit on the desktop model is rupees 4,500 and on portable model is rupees 5,000. So, let's start the solution. According to the given question, a merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost rupees 25,000 and rupees 40,000 respectively. Now, we have to determine the number of units of each type of computers which the merchant should stop to get maximum profit. Number of units, desktop model, a portable model, stopped, be x and y respectively. So, obviously we have x greater than equal to 0 and y greater than equal to 0. Now, we are given the cost of a desktop model is equal to rupees 25,000 and the cost of a portable model is equal to rupees 40,000. The merchant does not want to invest more than rupees 70 lakhs. So, the merchant maximum investment is equal to rupees 70 lakhs. Therefore, we have 25,000 x plus 40,000 y less than equal to 70 lakhs cost of x desktop model plus the cost of y portable model should be less than or equal to rupees 70 lakhs as a merchant does not want to more than rupees 70 lakhs. Again, we are given the total monthly demand of computers will not exceed 250 units since the total monthly demand of computers will not exceed 250 units. We have x plus y less than equal to 250, the number of units of desktop model stopped plus the number of units of portable model stopped should be less than equal to 250. Again, we are given the merchant's profit on the desktop model is rupees 4500 and on portable model is rupees 5000. So, the total profit in rupees is equal to 4500 x plus 5,005 lakh, z is equal to 4500 x plus 5,005 how the merchant wants to maximize his profit, is profit so the problem reduces to maximize z is equal to 4500 x plus 5000 y subject to the constraints 25000 x plus 40000 y less than equal to 70 lakhs 5x plus 8y less than equal to 1400 let us give this as number 1 so this is our first constraint x plus y less than equal to 250 let us give this as number 2 x greater than equal to 0 and y greater than equal to 0 let us give this as number 3 so z is equal to 4500 x plus 5000 y is our objective function we have to maximize that subject to the constraints 5x plus 8y less than equal to 1400 x plus y less than equal to 250 x greater than equal to 0 and y greater than equal to 0 now we will draw the graph and find the feasible region subject to these constraints so first we will draw the line corresponding to the inequality 5x plus 8y less than equal to 1400 now the equation corresponding to the inequality 5x plus 8y less than equal to 1400 is 5x plus 8y is equal to 1400 now when x is equal to 0 y is equal to 175 when y is equal to 0 x is equal to 280 so the points 0 175 and 280 0 satisfy the equation 5x plus 8 y is equal to 1400 so the graph of the line 5x plus 8 y is equal to 1400 can be drawn by plotting the points 0 175 and 280 and then joining them let a is the point 0 175 and b is the point 280 so a b represents the equation 5x plus 8 y is equal to 1400 now a b divides the plane into two half planes so we will consider the half plane which will satisfy one now clearly the origin satisfy this inequality so the half plane containing the origin is a graph of 1 again the equation corresponding to the inequality x plus y less than equal to 250 is x plus 5 is equal to 250 now clearly the points 0 250 and 250 0 satisfy the equation x plus 5 is equal to 250 so we will plot these points on the same graph to get the graph of the line x plus 5 is equal to 250 now let us take c as a point 0 250 and d as a point 250 0 so cd represents the equation of the line x plus y is equal to 250 again this line divides the plane into two half planes so we will consider the half plane which will satisfy to clearly the origin satisfy this inequality so the half plane containing the origin is a graph of 2 again x greater than equal to 0 and 5 greater than equal to 0 implies that the graph lies in the first quadrant only so now the green shaded region in this graph is the feasible region satisfying all the given constraints now clearly the shaded region is bounded so now we will find out the coordinates of the corner points of this feasible region now the coordinates of point a are 0 175 now the coordinates of the origin are 0 0 again the coordinates of the point tr 250 now let us take this point which is a point of intersection of the line 5 x plus 8 y is equal to 1400 and x plus y is equal to 250 as a point p now from this graph we observe that the coordinates of p are 250 so here the feasible region is bounded with coordinates of its corner points as a whose coordinates are 0 175 or whose coordinates are 00 b whose coordinates are 250 0 and p whose coordinates are 250 now according to the corner point method maximum value of set will occur at any of these points so we will evaluate z which is equal to 4500 x plus 5000 at each point now at the point 0 175 z is equal to 4500 x 0 plus 5000 x 175 and this is equal to 75000 origin z is equal to 4500 into 0 plus 5000 into 0 which is equal to 0 now again at the point 250 0 z is equal to 4500 into 250 plus 5000 into 0 and this is equal to 11 x 25000 now at 250 z is equal to 4500 into 200 plus 5000 into 50 and this is equal to 9,000 this is again equal to 11,000 50,000 so here we observe that the maximum value of z is equal to 11 like 50,000 which occurs when x is equal to 200 and y is equal to 50 therefore for maximum profit of rupees 11 like 50,000 the merchant should store 200 units desktop model three units portable model hence the answer for the above question is the merchant should store 200 units of desktop model and 50 units of portable model then his profit will be maximum and then the maximum profit will be rupees 11,50,000 so this completes our session I hope the solution is clear to you bye and have a nice day