 Good morning friends. I'm poor wine 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 potash and at most 310 kilograms of chlorine if the grower wants to maximize the amount of nitrogen added to the garden how many Bags of each brand should be added. What is the maximum amount of nitrogen added? 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 maximize the amount of nitrogen added to the garden How many bags of each brand should be added? So let the number of bags of brand P fertilizer BX and number of bags of brand Q fertilizer BY Now we are given the following table showing the amount of nitrogen Phosphoric acid potassium chlorine in a bag of each brand It is also given that the garden needs at least 240 kilograms of phosphoric acid at least 270 kilograms of potash and at most 310 kilograms of chlorine therefore According to the given conditions We have the following inequalities Now from the table we can clearly see that brand P contains 3 kilograms of nitrogen 1 kilogram of Phosphoric acid 3 kilograms of potash 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 potash and 2 kilograms of chlorine per bag Now 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 That's the total amount is X Plus 2 Y and this is greater than equal to 240 Because we are given that the garden needs at least 240 kilograms of phosphoric acid So this is our 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 That's the total amount is 3 X plus 1.5 Y 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 2 X plus Y is greater than equal to 180 And this is our 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 the total amount is 1.5 X plus 2 Y 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 3 X plus 4 Y is less than equal to 620. This is our 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 maximized. Therefore we get the objective function as Z is equal to 3 X plus 3.5 Y Now as number of bags of brand P fertilizer is X and each bag contains 3 kilograms of nitrogen So X bags will contain 3 X 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 3 X plus 3.5 Y and therefore the objective function becomes Z is equal to 3 X plus 3.5 Y Thus the linear programming problem becomes maximized Z is equal to 3 X plus 3.5 Y Subject to the constraints X plus 2 Y greater than equal to 240 2 X plus Y greater than equal to 180 3 X plus 4 Y 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 2 Y greater than equal to 240 and the line corresponding to this inequality is X plus 2 Y equal to 240 and the points 0 120 and 240 0 lie on this line The second inequality is 2 X plus Y greater than equal to 180 And the line corresponding to this inequality is 2 X plus Y equal to 180 and the points 0 180 and 90 0 lie on this line The third inequality is 3 X plus 4 Y less than equal to 620 and the line corresponding to this inequality is 3 X plus 4 Y equal to 620 And the points 0 155 and 620 upon 3 0 lie on this line Now we have plotted the points 0 120 and 240 0 satisfying the equation X plus 2 Y equal to 240 on the graph join them and we have named the line as AB We have plotted the points 0 180 and 90 0 satisfying the equation 2 X plus Y equal to 180 on the same graph Join them and we have named the line as CD And we have also plotted the points 0 155 and 620 upon 3 0 satisfying the equation 3 X plus 4 Y equal to 620 on the graph join them and we have named the line as EF The lines AB CD and EF intersect each other at the points L M and N 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 Thus the shaded portion in the graph is the feasible region satisfying all the given constraints Here the feasible region is triangle L M 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 the maximum value of Z will occur at Any of the above points? Therefore, we will calculate the value of Z is equal to 3 X plus 3.5 Y at these points Now Z is equal to 3 into 20 plus 3.5 into 140 at 21 40 and This comes out to be equal to 550 Z is equal to 3 into 40 plus 3.5 into 100 at 40 hundred and This comes out to be equal to 470 and Z is equal to 3 into 140 plus 3.5 into 50 at 140 50 and This comes out to be equal to 595 Hence the maximum value of Z out of these three values is 595 which occurs at 140 50 or we can say amount of nitrogen is maximum when X is equal to 140 and Y is equal to 50 Thus maximum amount of nitrogen is 595 kilograms when 140 bags of brand P and 50 bags of brand Q are used Thus we write our answer as 140 bags of brand P and 50 bags of brand Q and maximum amount of nitrogen is equal to 595 kilograms This is our answer. Hope you have understood the solution. Bye and take care