 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says there are two types of fertilizers F1 and F2. F1 consists of 10% of nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for our crop. If F1 cost rupees 6 per kg and F2 cost rupees 5 per kg determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost? Let us now start with the solution. First upon let us assume that let the quantities of fertilizers F1 and F2 used be XKG and YKG so we can write let quantities of fertilizers F1 and F2 used be XKG and YKG. Now clearly we can see F1 consists of 10% of nitrogen and F2 consists of 5% of nitrogen and we also know that farmer needs at least 14 kg of nitrogen. Now according to the given question 10 upon 100 X plus 5 upon 100 Y is greater than equal to 14. We know quantity of nitrogen in XKG of F1 is equal to 10% of X that is 10 upon 100 X and quantity of nitrogen in YKG of F2 is equal to 5% of Y that is 5 upon 100 Y. We know minimum requirement of nitrogen is equal to 14 kg so sum of these two quantities must be greater than equal to 14. Now multiplying both the sides of this in equation by 100 we get 10 X plus 5 Y is greater than equal to 1400. Now dividing both the sides of this in equation by 5 we get 2X plus Y is greater than equal to 280. This is our constraint 1. Now we are given F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid and minimum requirement of phosphoric acid is 14 kg. Now we get according to the question 6 upon 100 X plus 10 upon 100 Y is greater than equal to 14. We know quantity of phosphoric acid in XKG of F1 is equal to 6% of X that is 6 upon 100 X and quantity of phosphoric acid in YKG of F2 is equal to 10% of Y that is 10 upon 105. Now minimum requirement of phosphoric acid is equal to 14 kg so sum of these two quantities must be greater than equal to 14. Now multiplying both the sides of this in equation by 100 we get 6X plus 10 Y is greater than equal to 1400. Now dividing both the sides of this in equation by 2 we get 3X plus 5 Y is greater than equal to 700. Now this is our constraint 2. Now we know quantities of F1 and F2 required will be greater than or equal to 0. So we get X is greater than equal to 0 and Y is greater than equal to 0. This is our constraint 3 and this is our constraint 4. Here we know that X is the quantity of F1 and Y is the quantity of F2. Now we are given that F1 cost rupees 6 per kg and F2 cost rupees 5 per kg and we have to minimize the cost. Now we get objective function Z is equal to 6X plus 5 Y cost of 1 kg of F1 is equal to rupees 6 so cost of X kg of F1 is equal to 6X. We know cost of 1 kg of F2 is equal to rupees 5 so cost of Y kg of F2 is equal to 5 multiplied by Y that is 5 Y. Now total cost is equal to 6X plus 5 Y and it is equal to objective function Z. Now the required linear programming problem becomes minimize Z is equal to 6X plus 5 Y subject to constraints 2X plus Y is greater than equal to 280 3X plus 5 Y is greater than equal to 700 X is greater than equal to 0 and Y is greater than equal to 0. Now we will solve this linear programming problem graphically. Now clearly we can see points 0.280 and 140.0 lie on the line 2X plus Y is equal to 280. Now we will plot these two points on the graph and then we will draw this line by joining these two points. Now clearly we can see this point represents 0.280 and this point represents 140.0 joining these two points we get the line 2X plus Y is equal to 280. Now this line divides the plane into two half planes. We will consider the plane that satisfies the inequality 2X plus Y is greater than 280 that is the plane that does not contains the origin 00. Now we will draw a line 3X plus 5 Y is equal to 700 corresponding to this inequality. Now clearly we can see points 0.140 and 700 upon 3.0 lie on the line 3X plus 5 Y is equal to 700. Now we will plot these two points on the same graph and then we will join these two points to obtain this line. Now clearly we can see this point represents 0.140 this point represents 700 upon 3.0. Now joining these two points we get the line 3X plus 5 Y is equal to 700. Now this line divides the plane into two half planes. Now we will consider the plane that satisfies the inequality 3X plus 5 Y is greater than 700 that is the plane that does not contain the origin 00. Now next two constraints are X is greater than equal to 0 and Y is greater than equal to 0. This implies that the graph lies in the first quadrant only. Now this shaded region in the graph is the feasible region that satisfies all the given constraints. Clearly we can see feasible region is unbounded. Also note that these two lines that is 2X plus Y is equal to 280 and 3X plus 5 Y is equal to 700 intersect each other at point 100 comma 80. So coordinates of this point are 100 comma 80. Now we know if the minimum value of the unbounded feasible region exists it occurs at corner points of the feasible region. Now we know corner point of a feasible region is a point in the region which is the intersection of two boundary lines. So here for this unbounded feasible region corner points are 0 comma 280, 100 comma 80 and 700 upon 3 comma 0. Now we can write the shaded portion of the graph shows the feasible region satisfying all the given constraints. The feasible region is unbounded. Now corner points of the feasible region are 0 comma 280, 100 comma 80 and 700 upon 3 comma 0. Now we know according to the corner point method minimum value of objective function z will occur at any of these points. So we will calculate the value of z at all these points. Now we know objective function z is equal to 6X plus 5 Y. Now first of all we will find the value of z at corner point 0 comma 280. Now substituting 0 for X and 280 for Y in this expression we get 6 multiplied by 0 plus 5 multiplied by 280 is equal to z. Now this is further equal to 1400. So we can write z is equal to 1400 at point 0 comma 280. Similarly we can find value of z at this corner point that is 100 comma 80. Substituting 100 for X and 80 for Y in this expression we get 6 multiplied by 100 plus 5 multiplied by 80 which is further equal to 1000. So we get z is equal to 1000 at point 100 comma 80. Similarly, value of z at point 700 upon 3 comma 0 is equal to 1400. Now clearly we can see minimum value of z is equal to 1000 and it occurs at corner point 100 comma 80. Now we can write thus minimum value of z is equal to 1000 at corner point 100 comma 80 that is X is equal to 100 and Y is equal to 80. Now 1000 is the minimum value of the objective function z. If open half plane z less than 1000 does not have any point common with the feasible region. Now we will check if the region represented by z less than 1000 has any common point to the feasible region. Now we know z is less than 1000 implies 6x plus 5y is less than 1000. Now we will draw a line 6x plus 5y is equal to 1000 corresponding to this inequality on the graph. Now clearly we can see points 0 comma 200 and 100 comma 80 lie on the line 6x plus 5y is equal to 1000. Now this is the point 0 comma 200 and this is the point 100 comma 80 joining these two points we get the line 6x plus 5y is equal to 1000. Now this yellow shaded portion in the graph excluding this line represents 6x plus 5y is less than 1000 or we can say z is less than 1000. Clearly we can see this region has no point common with the feasible region. So we get z is not less than 1000 in the feasible region except at 100 comma 80. Now we can write the yellow shaded portion in the graph excluding line 6x plus 5y is equal to 1000 represents 6x plus 5y is less than 1000 or we can say it represents z is less than 1000. Now this region has no point common with the feasible region thus minimum value of z is equal to 1000 which occurs at corner point 100 comma 80 or we can say it occurs at x is equal to 100 and y is equal to 80. Now we know z represents the cost so the minimum cost is equal to rupees 1000. x represents the quantity of fertilizer f1 so quantity of fertilizer f1 is equal to 100 kg, y represents the quantity of fertilizer f2 so quantity of fertilizer f2 is equal to 80 kg. Now this is our required answer this completes the session hope you understood the solution take care and have a nice day.