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  • Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 23

Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 23

Answers (1)

Answer:

Minimum cost = Rs.92 when 100kg of type I fertilizer and 80 kg of Type II fertilizer is supplied.

Hint:

Form Linear Equation and solve graphically.

Given:

A gardener supply fertilizer of type I which consists of 10% of nitrogen and 6% phosphoric acid and Type II fertilizer which consist of 5% nitrogen and 10% of phosphoric acid. After testing the soil conditions, he finds that he needs at least 14 kg of nitrogen and 14kg of phosphoric acid for two crop of the type I fertilizer cost 60P/Kg and types II fertilizer 40P/Kg

Solution:

Let x kg of Type I fertilizer and y kg of Type II fertilizers are supplied.

The quantity of fertilizers cannot be negative.

So, x, y \geq 0

A gardener has a supply of fertilizer of type I which consists of 10% nitrogen and type II consists of 5% nitrogen, and he needs at least 14kg for his crop.

So,

                (10 \times 100)+(5 \times 100) \geq 14 \\

Or           \begin{aligned} & &10 x+5 y \geq 1400 \end{aligned}

A gardener has a supply of fertilizer of type I which consists of 6% phosphoric acid and type II consists of 10% phosphoric acid, and he needs at least 14 kg of phosphoric acid for his crop.

Sp,

                (6 \times 100)+(10 \times 100) \geq 14 \\

Or           \begin{aligned} & &6 x+10 y \geq 1400 \end{aligned}

Therefore, A/Q, constraint is,

                \begin{aligned} &10 x+5 y \geq 1400 \\ &6 x+10 y \geq 1400 \end{aligned}

If the type I fertilizer costs 60 paise per kg and Type II fertilizer costs 40 paise per kg. Therefore, the cost of x kg of Type 1 fertilizer and y kg of Type II fertilizer is Rs.0.60x and Rs.0.40y respectively.

Total cost=Z(let)=0.6x + 0.4y is to be minimized.

Thus the mathematical formulation of the given LPP ism,

                Min Z=0.6 + 0.4y

Subject to the constraints,

                \begin{aligned} &10 x+5 y \geq 1400 \\ &6 x+10 y \geq 1400 \\ &x, y \geq 0 \end{aligned}

Region represented by 6 x+10 y \geq 1400 : the line 6 x+10 y=1400  passes through  \mathrm{A}\left(\frac{700}{3}, 0\right) , B(0,140). The region which does not contains origin represents solutions of the in equation 6 x+10 y \geq 1400  as (0,0) doesn’t satisfy the in equation 6 x+10 y \geq 1400

Region represented by 10 x+5 y \geq 1400 : the line 10 x+5 y \geq 1400  passes through C(140,0) and D(0,280). The region which does not contains origin represents solutions of the in equation 10 x+5 y \geq 1400  as (0,0) doesn’t satisfy the in equation 10 x+5 y \geq 1400

The region  x, y \geq 0 : represents the first quadrant.

 The corner points are D(0,280),E(100,80), A(700/3,0)

The values of Z at these points are as follows:

                                 

Corner Points

z=0.6x+0.4y

O

0

D

112

E

92

F

140

The minimum value of Z is Rs.92 which is attained at E(100,80)

Thus, the minimum cost is Rs.92 obtained when 100kg of Type I fertilizer and 80 kg of Type II fertilizer is supplied.

 

Posted by

infoexpert27

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