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10. There are two types of fertilisers F1 and F2 . F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

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Let farmer buy x kg of fertilizer F1 and y kg of F2 .Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in table as :

	Nitrogen	phosphoric acid	Cost
F1	10	6	6
F2	5	10	5
requirement	14	14

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen

$10\%x+5\%y\geq 14$

$\frac{x}{10}+\frac{y}{20}\geq 14$

$2x+y\geq 280$

F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen

$6\%x+10\%y\geq 14$

$\frac{6x}{100}+\frac{y}{20}\geq 14$

$3x+56y\geq 700$

Total cost is Z . $Z=6x+5y$

Subject to constraint ,

$2x+y\geq 280$

$3x+56y\geq 700$

$x\geq 0,y\geq 0$

$Z=6x+5y$

The feasible region determined by constraints is as follows:

It can be seen that feasible region is unbounded.

The corner points of feasible region are $A(\frac{700}{3},0),B(100,80),C(0,280)$

The value of Z at corner points is as shown :

corner points	$Z=6x+5y$
$A(\frac{700}{3},0)$	1400
$,B(100,80)$	1000	minimum
$C(0,280)$	1400

Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw $6x+5y< 1000$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $6x+5y< 1000$ .

Hence , Z has minimum value 1000 at point $,B(100,80)$

Posted by

seema garhwal

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