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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 47

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 47

Answers (1)

Answer: Therefore, the minimum cost is Rs.1000

Hint:

Use properties of LPP

Given:

Two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid if F1 consists of Rs.6/kg and F2 costs Rs.5/Kg

Solution:

Suppose x kg of fertilizer F1 and y kg of fertilizer F2 is used to meet the nutrient requirements.

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. Bu the farmer needs at least 14 kg of nitrogen for the crops.

\begin{aligned} &10 \% \text { of } x \mathrm{~kg}+5 \% \text { of } y \geq 14 \mathrm{~kg} \\ &\Rightarrow \frac{x}{10}+\frac{y}{5} \geq 14 \\ &\Rightarrow 2 x+y \geq 280 \end{aligned} 

Similarly, F1 consists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. But the farmer need at least 14 kg of phosphoric acid for the crops.

6 \% \text { of } x \mathrm{~kg}+10 \% \text { of } y \geq 14 \mathrm{~kg} 

\begin{aligned} &\Rightarrow \frac{6 x}{10}+\frac{10 y}{100} \geq 14 \\ &\Rightarrow 3 x+5 y \geq 700 \end{aligned}

The cost of fertilizer F1 is Rs.6/kg and Fertilizer F2 in Rs.5/Kg, therefore total cost of x kg of fertilizer F1 and y kg of fertilizer F2 is Rs.(6x+5y)

Thus, the given linear programming problem is

Minimize Z = 6x + 5y

Subject to the constraints

                                 \begin{aligned} &2 x+y \geq 280 \\ &3 x+5 y \geq 700, x, y \geq 0 \end{aligned}

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

The coordinates of the corner points of the feasible region are

    \mathrm{A}\left(\frac{700}{3}, 0\right), \mathrm{B}(100,80) \& \mathrm{C}(0,280)

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

 

Corner Points

Z=6x+5y

A\left(\frac{700}{3}, 0\right)

6 \times \frac{700}{3}+5 \times 0=1400

\mathrm{B}(100,80)

6 \times 100+5 \times 80=1000 (minimum)

C(0,280)

6 \times 0+5 \times 280=1400

 

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

It can be sure 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 F1 and 80 kg of fertilizer F2 should be used so that the nutrient requirements are met at minimum cost.

The minimum cost is Rs.1000

 

Posted by

infoexpert27

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