You visited us 1 times! Enjoying our articles? Unlock Full Access!

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?

Open in App

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.

Suggest Corrections

Similar questions

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 ?

Q. There are two types of fertilisers

F_{1}

and

F_{2}

F_{1}

consists of

10 %

nitrogen and

6 %

phosphoric acid,

F_{2}

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

F_{1}

costs Rs

6 / kg

and

F_{2}

costs Rs

5 / kg

, determine how much amount of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Q. There are two types of fertilisers

F_{1}

and

F_{2}

F_{1}

consists of

10 %

nitrogen and

6 %

phosphoric acid,

F_{2}

consists of

5 %

nitrogen and

10 %

phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least

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

, then the minimum cost of mixture which will produce the required nutrients is

Q. A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6% phosphoric acid and type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg, determine how many kilograms of each fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Q. There are two types of fertilisers 'A' and 'B' . 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If 'A' costs ₹10 per kg and 'B' cost ₹8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost

Join BYJU'S Learning Program

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