Graphical Method of Solving Linear Programming Problems
If the grower...
Question
If the grower wants to maximise 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?
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Solution
Let's assume that bags of brand P needed be X and brand Q be Y
Since bag of brand P contains 1 kg and bag of brand Q contains 2 kg of Phosphoric acid. Also, garden needs at least 240 kg of phosphoric acid.
∴X+2Y≥240...(1)
Since bag of brand P contains 3 kg and bag of brand Q contains 1.5 kg of potash. Also, garden needs at least 270 kg of potash.
∴3X+1.5Y≥270
⇒2X+Y≥180...(2)
Since bag of brand P contains 1.5 kg and bag of brand Q contains 2 kg of chlorine. Also, garden needs at most 310 kg of chlorine.
∴1.5X+2Y≤310...(3)
Since count of bags can't be negative.
∴X≥0,Y≥0...(4)
Now, Bag of brand P contains 3 kg of Nitrogen and Bag of brand Q contains 3.5 kg of Nitrogen.
So, total Nitrogen contents Z=3X+3.5Y
We have to maximise the amount of Nitrogen added to the garden.
After plotting all the constraints given by equation (1), (2), (3) and (4) we get the feasible region as shown in the image.
Corner points
Value of Z=3X+3.5Y
A (20, 140)
550
B (40, 100)
470
C (140, 50)
595 (maximum)
Hence, to maximise the amount of nitrogen added to the garden, grower should use 140 bags of brand P and 50 bags of brand Q. In this case maximum amount of nitrogen added to the garden will be 595Kg