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

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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

 

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