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Forming a linear equation

There are two...

Question

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?

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Solution

Suppose x kg of fertilizer F₁and and y kg of fertilizer F₂ is used to meet the nutrient requirements.

F₁consists of 10% nitrogen and F₂consists of 5% nitrogen. But, the farmer needs atleast 14 kg of nitorgen for the crops.

∴ 10% of x kg + 5% of y kg ≥ 14 kg

$\Rightarrow \frac{x}{10} + \frac{y}{20} \geq 14 \Rightarrow 2 x + y \geq 280$

Similarly, F₁consists of 6% phosphoric acid and F₂consists of 10% phosphoric acid. But, the farmer needs atleast 14 kg of phosphoric acid for the crops.

∴ 6% of x kg + 10% of y kg ≥ 14 kg

$\Rightarrow \frac{6 x}{100} + \frac{10 y}{100} \geq 14 \Rightarrow 3 x + 5 y \geq 700$

The cost of fertilizer F₁is ₹6/kg and fertilizer F₂is ₹5/kg, therefore, total cost of x kg of fertilizer F₁and and y kg of fertilizer F₂is ₹(6x + 5y).

Thus, the given linear programming problem is

Minimise Z = 6x + 5y

subject to the constraints

2x + y ≥ 280

3x + 5y ≥ 700

x, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,

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

The value of the objective function at these points are given in the following table.

Corner Point Z = 6x + 5y

$(\frac{700}{3}, 0)$ $6 \times \frac{700}{3} + 5 \times 0 = 1400$

(100, 80) 6 × 100 + 5 × 80 = 1000 → Minimum

(0, 280) 6 × 0 + 5 × 280 = 1400

The smallest value of Z is 1000 which is obtained at x = 100, y = 80.

It can be seen that the open half-plane represented by 6x + 5y < 1000 has no common points with the feasible region.

So, the minimum value of Z is 1000.

Hence, 100 kg of fertilizer F₁and and 80 kg of fertilizer F₂ should be used so that the nutrient requirements are met at minimum cost. The minimum cost is ₹1,000.

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Corner Point	Z = 6x + 5y
$(\frac{700}{3}, 0)$	$6 \times \frac{700}{3} + 5 \times 0 = 1400$
(100, 80)	6 × 100 + 5 × 80 = 1000 → Minimum
(0, 280)	6 × 0 + 5 × 280 = 1400