 Good morning friends, I am Purva and today we will work out the following question. A fruit grower can use two types of fertilizer in his garden, Brand P and Brand Q. The amounts of nitrogen, phosphoric acid, potassium, chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kilograms of phosphoric acid, at least 270 kilograms of potassium and at most 310 kilograms of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden, how many? Bags of each brand should be used. What is the minimum amount of nitrogen added in the garden? And this is the table given to us. 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 have, if the grower wants to minimize the amount of nitrogen added to the garden, how many bags of each brand should be used? So, let the number of bags of Brand P fertilizer, BX and number of bags of Brand Q fertilizer be Y. Now we are given the following table showing the amount of nitrogen, phosphoric acid, potassium and chlorine in a bag of each brand. We are also given that the garden needs at least 240 kilograms of phosphoric acid, at least 270 kilograms of potassium and at most 310 kilograms of chlorine. Therefore, according to the given conditions, we have the following inequalities. Now from this table we can clearly see that Brand P contains 3 kilograms of nitrogen, 1 kilogram of phosphoric acid, 3 kilograms of potassium and 1.5 kilograms of chlorine per bag. And Brand Q contains 3.5 kilograms of nitrogen, 2 kilograms of phosphoric acid, 1.5 kilograms of potassium and 2 kilograms of chlorine per bag. Now we have the number of bags of Brand P fertilizer is X and each bag contains 1 kilogram of phosphoric acid so X bags will contain X kilograms of phosphoric acid and number of bags of Brand Q fertilizer is Y and each bag contains 2 kilograms of phosphoric acid so Y bags will contain 2 into Y kilograms of phosphoric acid. So we get the total amount of phosphoric acid as X plus 2Y and this should be greater than equal to 240 because we are given that the garden needs at least 240 kilograms of phosphoric acid. So this is a constraint for phosphoric acid. Now again number of bags of Brand P fertilizer is X and each bag contains 3 kilograms of potash so X bags will contain 3 into X kilograms of potash and number of bags of Brand Q fertilizer is Y and each bag contains 1.5 kilograms of potash so Y bags will contain 1.5 into Y kilograms of potash. So the total amount is 3X plus 1.5Y and this is greater than equal to 270 because we are given that the garden needs at least 270 kilograms of potash. Now this implies 2X plus Y is greater than equal to 180 so this is a constraint for potash. Now again number of bags of Brand P fertilizer is X and each bag contains 1.5 kilograms of chlorine so X bags will contain 1.5 into X kilograms of chlorine and number of bags of Brand Q fertilizer is Y and each bag contains 2 kilograms of chlorine so Y bags will contain 2 into Y kilograms of chlorine. So we get the total amount as 1.5X plus 2Y and this is less than equal to 310 because we are given that the garden needs at most 310 kilograms of chlorine. Now this implies 3X plus 4Y is less than equal to 620 so this is a constraint for chlorine. Also we have X greater than equal to 0 and Y greater than equal to 0 as the number of bags of fertilizer used is greater than equal to 0. Now as the amount of nitrogen in the bags of Brand P and Brand Q fertilizer is 3 kilograms and 3.5 kilograms per bag respectively and it is to be minimized. Therefore we get the objective function as Z is equal to 3X plus 3.5Y. Now number of bags of Brand P fertilizer is X and each bag contains 3 kilograms of nitrogen so X bags will contain 3X kilograms of nitrogen and number of bags of Brand Q fertilizer is Y and each bag contains 3.5 kilograms of nitrogen so Y bags will contain 3.5 into Y kilograms of nitrogen. So the total amount is 3X plus 3.5Y and therefore we get the objective function as Z is equal to 3X plus 3.5Y. Thus the linear programming problem becomes minimize Z is equal to 3X plus 3.5Y subject to the constraints X plus 2Y greater than equal to 240, 2X plus Y greater than equal to 180, 3X plus 4Y less than equal to 620, X greater than equal to 0 and Y greater than equal to 0. Now the first inequality is X plus 2Y greater than equal to 240 and the line corresponding to this inequality is X plus 2Y equal to 240 and the points 0, 120 and 240, 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 X plus 2Y equal to 240 corresponding to the inequality X plus 2Y greater than equal to 240 on the graph by plotting the points 0, 120 and 240, 0 and then joining them. So we have plotted the points 0, 120 and 240, 0 on the graph joined 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 X plus 2Y greater than 240 that is the portion of the graph about this line AB along with the line AB in the feasible region for this inequality. Now the second inequality is 2X plus Y greater than equal to 180 and the line corresponding to this inequality is 2X plus Y equal to 180 and the points 0, 180 and 90, 0 lie on this line. So now we shall represent 2X plus Y greater than equal to 180 graphically by drawing the line 2X plus Y equal to 180 by plotting the points 0, 180 and 90, 0 satisfying the equation and then joining them. So we have plotted the points 0, 180 and 90, 0 on the graph joined them and we have named the line as CD. This line CD divides the plane into two half planes, we shall take the half plane that satisfies 2X plus Y greater than 180 that is the portion of the graph to the right-hand side of this line CD along with the line CD in the feasible region for this inequality. Now the third inequality is 3X plus 4Y less than equal to 620 and the line corresponding to this inequality is 3X plus 4Y equal to 620 and the points 0, 155 and 620 upon 3, 0 lie on this line. So now we shall represent 3X plus 4Y less than equal to 620 graphically by drawing the line 3X plus 4Y equal to 620 by plotting the points 0, 155 and 620 upon 3, 0 satisfying the equation and then joining them. So we have plotted the points 0, 155 and 620 upon 3, 0 on the same graph joined them and we have named the line as EF. This line EF divides the plane into two half planes, we shall take the half plane satisfying 3X plus 4Y less than 620 that is the portion of the graph below this line EF along with the line EF in the feasible region for this inequality. Also we have X greater than equal to 0 and Y greater than equal to 0 which implies that the graph lies in the first quadrant only. The lines A, B, CD and EF intersect at the 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 triangle L, M and N with coordinates of vertices as 21, 40, 40, 100 and 140, 50. 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, minimum value of Z will occur at any of the above points. Therefore we will calculate the value of Z is equal to 3X plus 3.5Y at these points. Now Z is equal to 3 into 20 plus 3.5 into 140 at 2140 and this comes out to be equal to 550. Z is equal to 3 into 40 plus 3.5 into 100 at 400 and this comes out to be equal to 470. And Z is equal to 3 into 140 plus 3.5 into 50 at 14050 and this comes out to be equal to 595. Hence the minimum value of Z out of these three values is 470 which occurs at 400 or we can say amount of nitrogen is minimum when X is equal to 40 and Y is equal to 100. Thus minimum amount of nitrogen is 470 kilograms when 40 bags of brand P and 100 bags of brand Q are used. Thus we write our answer as 40 bags of brand P and 100 bags of brand Q and minimum amount of nitrogen is equal to 470 kilograms. This is our answer. Hope you have understood the solution. Bye and take care.