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  • Please Solve RD Sharma Class 12 Chapter 29 Linner Programming Exercise 29.4 Question 1 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 29 Linner Programming Exercise 29.4 Question 1 Maths Textbook Solution.

Answers (1)

Answer:

Distance travelled at speed of  25 Km/hr=\frac{50}{3} Km  and at a speed of  40 Km/hr=\frac{40}{3}Km

Hint:

Time\: taken=\frac{Distance}{Speed}

Given:

Man spend  Rs\; 2/ Km \; on \; p\left | x_{0} \right |   when speed is 23km/hr and Rs 5/km when speed Rs 50 Km/hr.

Solution:

Let he drives x Km at sped of 25 Km/hr. and y Km/hr. at a speed of 40 Km/hr.

Let Z be total distance travelled by him so,

z=x+y

Since he spend Rs 2 per Km on petrol when speed is 25 Km/hr. and Rs5 per Km on petrol when speed is 40 Km/hr, So, expense on x Km and y Km or Rs2x and Rs2y respectively. But he has only Rs100, 50.

2x+5y≤100 (1st constraint)

Time\; taken\; to \; travel\; x\; Km=\frac{Distance}{Speed}

= \frac{x}{25}hr

Time taken to travel  y\: Km=\frac{y}{40}hr

Given he has 1 hr. to travel, so,
\begin{aligned} &\frac{x}{25}+\frac{y}{40} \leq 1 \\ & \end{aligned}

\Rightarrow 40 x+25 y \leq 1000 \\

\Rightarrow 8 x+5 y \leq 200(\text { second constraint })

Hence, mathematical formulation of LPP is find x and y which

Maximum  z=x+y

Subject to constraint

\begin{aligned} &2 x+5 y \leq 100 \\ & \end{aligned}

8 x+5 y \leq 200 \\

x, y \geq 0   since distance cannot be less than zero

The corresponding line is

\begin{aligned} &\Rightarrow 2 x+5 y \leq 100 \\ & \end{aligned}

\Rightarrow \frac{x}{50}+\frac{y}{20}=1       ....(i)

And\: 8 x+5 y=200

\Rightarrow \frac{x}{25}+\frac{y}{40}=1       …(ii)

Point P\left(\frac{50}{3}, \frac{40}{3}\right)   is obtain by solving                               

\begin{aligned} &8 x+5 y=200 \\ &2 x+5 y=100 \end{aligned}                                                           

 

Points

z=x+y   

O\left ( 0,0 \right )

 0

A _{2}\left ( 25,0 \right )

 25

P\left(\frac{50}{3}, \frac{40}{3}\right)

 30

B_{1}\left ( 0,20 \right )

20

 

And  z=30\: at \: z=\frac{50}{3}, z=\frac{40}{3}

Distance travelled at speed of  25 Km/hr=\frac{50}{3} Km and at a speed of  40 Km/hr=\frac{40}{3}Km

Maximum distance =30 Km

Posted by

infoexpert27

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