There are two types of fertilisers $F_{1}$ and $F_{2} \cdot F_{1}$ consists of $10$ % nitrogen and $6$ %phosphoric acid and $F_{2}$ consists of $5$ % nitrogen and $10$ % phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast $14 k g$ of nitrogen and $14 k g$ of phosphoric acid for her crop. If $F_{1}$ costs $R s . 6 / k g$ and $F_{2}$ costs $R s . 5 / k g$ , 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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Solution

Let's assume that amount of fertiliser $F_{1}$ used be $X$ kg and amount of fertiliser $F_{2}$ used be $Y$ kg.

Since, $F_{1}$ consist $10 %$ and $F_{2}$ consist $5 %$ nitrogen. Also, farmer needs at least 14 kg of nitrogen.

$∴ (10 % o f X) + (5 % o f Y) \geq 14$

$\Rightarrow \frac{10 X}{100} + \frac{5 Y}{100} \geq 14$

$\Rightarrow 2 X + Y \geq 280 . . . (1)$

Again since, $F_{1}$ consist $6 %$ and $F_{2}$ consist $10 %$ phosphoric acid. Also, farmer needs at least 14 kg of phosphoric acid.

$∴ (6 % o f X) + (10 % o f Y) \geq 14$

$\Rightarrow \frac{6 X}{100} + \frac{10 Y}{100} \geq 14$

$\Rightarrow 3 X + 5 Y \geq 700 . . . (2)$

Since amount of fertilisers can't be negative.

$∴ X \geq 0, Y \geq 0 . . . (3)$

Since, $F_{1}$ costs $6$ Rs per kg and $F_{2}$ costs $5$ Rs per kg.

So, total costs $(Z) = 6 X + 5 Y$

We have to minimise total costs of farmer subjects to the constraints given by (1), (2) and (3).

After plotting all the constraints given by equation (1), (2) and (3) we got the feasible region as shown in the image.

Corner points Value of $Z = 6 X + 5 Y$
A $(0, 180)$ $1400$
B $(100, 80)$ $1000$ (minimum)
C $(\frac{700}{3}, 0)$ $1400$
Now, since region is unbounded. Hence we need to confirm that minimum value obtained through corner points is true or not.

Now, after plotting the region $Z < 1000$ to check if there exist some points in feasible region where value can be less than 1000.

$\Rightarrow 6 X + 5 Y < 1000$

Since there is no other points than B are common points between the feasible region and the region which contains $Z < 1000$ . So 1000 is the minimum cost.

Hence, for minimum costs, $F_{1}$ should be $100$ kg and $F_{2}$ should be $80$ kg. In this case, minimum costs to farmer will be $1000 R s$

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