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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Solution

x kg of fertiliser F₁ and y kg of F₂ fertiliser are supplied.
Quantity of fertiliser cannot be negative.
Therefore, $x, y \geq 0$

A gardener has supply of fertiliser F₁ which consists of 10% nitrogen and fertiliser F₂ consists of 5% nitrogen and he needs at least 14 kg of nitrogen for his crop.
$\frac{10 x}{100} + \frac{5 y}{100} \geq 14 \Rightarrow 10 x + 5 y \geq 1400$

A gardener has supply of fertiliser F₁ which consists 6% phosphoric acid and fertiliser F₂ consists of 10% phosphoric acid.
And he needs 14 kg of phosphoric acid for his crop.

$\frac{6 x}{100} + \frac{10 y}{100} \geq 14 \Rightarrow 6 x + 10 y \geq 1400$

Therefore, according to the question, constraints are
$10 x + 5 y \geq 1400 6 x + 10 y \geq 1400$

If F₁ costs Rs 6/kg and F₂ costs Rs 5/kg.Therefore, cost of x kg of fertiliser F₁ and y kg of fertiliser F₂ is Rs 6x and Rs 5y respectively.

Total cost = Z = $Rs (6 x + 5 y)$ which is to be minimised.

Thus, the mathematical formulation of the given linear programmimg problem is

Min Z = $6 x + 5 y$

subject to

$6 x + 10 y \geq 1400 10 x + 5 y \geq 1400$
$x, y \geq 0$

First we will convert inequations into equations as follows:
6x + 10y = 1400, 10x + 5y = 1400, x = 0 and y = 0

Region represented by 6x + 10y ≥ 1400:
The line 6x + 10y = 1400 meets the coordinate axes at $A (\frac{700}{3}, 0)$ and B(0, 140) respectively. By joining these points we obtain the line
6x + 10y = 1400. Clearly (0,0) does not satisfies the 6x + 10y = 1400. So, the region which does not contain the origin represents the solution set of the inequation 6x + 10y ≥1400.

Region represented by 10x + 5y ≥ 1400:
The line 10x + 5y = 1400 meets the coordinate axes at C(140, 0) and $D (0, 280)$ respectively. By joining these points we obtain the line
10x + 5y = 1400. Clearly (0,0) does not satisfies the inequation 10x + 5y ≥ 1400. So,the region which does not contain the origin represents the solution set of the inequation 10x + 5y ≥ 1400.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 6x + 10y ≥1400, 10x + 5y ≥ 1400, x ≥ 0 and y ≥ 0 are as follows.

The corner points are D(0, 280), E(100, 80) and $A (\frac{700}{3}, 0)$ .

The values of Z at these corner points are as follows

Corner point Z = $6 x + 5 y$

D 1400

E 1000

A 1400

The minimum value of Z is Rs 1000 which is attained at E(100, 80).
Thus, the minimum cost is Rs 1000 obtained when 100 kg of fertiliser F₁ and 80 kg of fertiliser F₂ were supplied.

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Q. There are two types of fertilizers F₁and F₂. F₁consists of 10% nitrogen and 6% phosphoric acid and F₂consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F₁costs ₹6/kg and F₂costs ₹5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?

There are two types of fertilizers $F_{1}$ and $F_{2}$ . $F_{1}$ consists of 10 % nitrogen and 6 % phosphoric acid and $F_{2}$ consists of 5 % nitrogen and 10 % of 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 $F_{1}$ costs Rs. 6 kg and $F_{2}$ costs Rs. 5 kg, determine how much of each at a minimum cost. What is the minimum cost ?