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  • Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 54

Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 54

Answers (1)

Answer: The total minimum cost of the fertilizers is Rs.1980

Hint:

 Use property of LPP

Given:

A consists of 12% of nitrogen and 5% of phosphoric acid at costs Rs.10/kg and B consists of 4% of nitrogen and 5% of phosphoric acid at costs Rs.8/kg

Solution:

The given information can be tabulated as follows.

 

Fertilizer

Nitrogen

Phosphoric Acid

Cost/Kg

A

12%

5%

10

B

4%

5%

8

 

Let the requirement of fertilizer A by the farmer be x kg and that of B be y kg

It is given that farmer requires at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops.

The in equations thus formed based on the given information are as follows.

                    \begin{aligned} &\frac{12}{100} x+\frac{4}{100} y \geq 12 \\ &\Rightarrow 12 x+4 y \geq 1200 \\ &\Rightarrow 3 x+y \geq 300 \end{aligned}

Also,

                 \begin{aligned} &\frac{5}{100} x+\frac{5}{100} y \geq 12 \\ &\Rightarrow 5 x+5 y \geq 1200 \\ &\Rightarrow x+y \geq 240 \end{aligned}

Total cost of the fertilizer Z = Rs.(10x+8y)

Therefore, the mathematical formulation of the given LPP is

Minimize Z=10x+8y

Subject to the constraints

                  3 x+y \geq 300 \\                                                                                  … (i)

                 x+y \geq 240 \\                                                                                     … (ii)

                 \begin{aligned} & &x \geq 0, y \geq 0 \end{aligned}                                                                                    … (iii)

The feasible region determined by constraints (1) to (3) is graphically represented as

X+Y=240 \\

3 X+Y=300 \quad 10 X+8 Y=1980

Hence, it is seen that the feasible region is unbounded. The value of Z at the corner points of the feasible region are represented in tabular form as

 

Corner Points

z=10x+8y

(0,300)

10 \times 0+8 \times 300=2400

(30,210)

10 \times 30+8 \times 210=1980 (minimum)

B(240,0)

 10 \times 2400+8 \times 10=2400

The open half plane determined by  10 x+8 y \leq 1980   has no point in common with the feasible region. So, the minimum value of Z is 1980.

The minimum value of Z is 1980, which is obtained at x = 30 and y = 210.

Thus, the minimum requirement of fertilizer of type A will be 30kg and that of type B will be 210 kg.

Also, the total minimum cost of the fertilizers is Rs.1980.

 

Posted by

infoexpert27

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