A fruit grower can use two types of fertiliser in his garden, brand p and brand q. The amounts ( in kg ) of nitrogen, phosphonic acid, Potash,and Chlorine in a bag of each brand are given in the table. Tests indicate
that the garden needs at least 240 kg of phosphonic acid, at least 270 kg of Potash and at most 310 kg of chlorine. If the grower wants to minimise the amount of nitrogen added to the garden, How many bags of each bread should be used? What is the minimum amount of nitrogen added in the garden? Kg per bag brand P brand Q nitrogen 3,3.5 phosphoric acid 1,2 potash 3,1.5 chlorine 1.5,2.

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P=140, Q=50, Min. Amount 545

P=20, Q=140, Min. Amount 550 kg

P=100, Q=40, Min. Amount 470 kg

P=40, Q=100, Min. Amount 470 kg

answer is D.

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Detailed Solution

Let, take bags of brand P needed= x Bag of brand Q needed = y
Given that bag of brand P contains 1 kg and bag of brand Q contains 2 kg of phosphoric acid.
So, x+2y≥ 240 - (i) (∵ Garden needs at least 240 kg phosphoric acid)
Bags of brand P contain 3 kg and bags of brand Q contain 1.5 kg of potash.
So, 3x+1.5y ≥ 270-(ii) (∵ Garden needs at least 270 kg potash)
Bag of brand P contains 1.5 kg and bag of Q contains 5.2 kg chlorine
So, 1.5x + 2 y ≤ 310 -(iii) (∵ Garden needs at least 310 kg chlorine)
Bag of brand p contains 3 kg of nitrogen and bag of brand Q contain 3.5 kg of Nitrogen So, total nitrogen = N= 3x+3.5y - (iv)
We know that quantity of bag required can’t be negative So, we get=x≥0 and y≥0 -(v)
Now we have to plot graph for all equations (i),(ii),(iii),(iv) and (v)
Question Image

From the graph we get that
At point A (20,40) Value of N = 3x+3.5y = 550
AT point B(40,100)= Value of N = 3x+3.5y = 470 (minimum) At point C(140,50) Value of N = 3x+3.5y = 545
Hence, to minimise the amount of nitrogen, growers should use 40 bags of brand P and 100 bags of brand Q. The minimum amount of nitrogen added will be 470 kg.

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