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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?


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.
$$\therefore X+2Y\geq 240\ \ \ ...(1)$$

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.
$$\therefore 3X+1.5Y\geq 270$$

$$\Rightarrow 2X+Y\geq 180\ \ \ ...(2)$$

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.
$$\therefore 1.5X+2Y\leq 310\ \ \ ...(3)$$

Since count of bags can't be negative.
$$\therefore X\geq 0, Y\geq 0\ \ \ ...(4)$$

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 $$595\ Kg$$

816504_847077_ans_84c80f17ff644e738f147425d0f27a43.jpg

Mathematics

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