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  • Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 36

Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 36

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Answer: The maximum that the man can travel in 1 hr. is 30km. Distance travelled at the speed of 25km/hr is \frac{50}{3} km and 40 km/hr. is \frac{40}{3} km

Hint:

Let us assume that the man travels xkm/hr. and y km when the speed is 40km/hr.

Given:

Speed of vehicle=25km/hr.

Per km cost on petrol=2Rs/km

Young man carries only =100 Rs.

Solution:                   

Let us assume that the man travels x km when the speed is 25km/hour and y km when the speed is 40km/hour.

Thus, the total distance travelled is (x+y) km.

Now, it’s given that the man has 100Rs to spend on petrol.

Total cost of petrol=  2 x+5 y \leq 100

Now, Time taken to travel x km = \frac{x}{25}  hour

Time taken to travel y km =  \frac{y}{40} hour

Now it’s given that the maximum time is 1 hour. So,

             \begin{gathered} \frac{x}{25}+\frac{y}{40} \leq 1 \\ \Rightarrow 8 x+5 y \leq 200 \end{gathered}

Thus, the given linear programming problem is

Maximize Z=x+y

Subject to the constraints

            \begin{aligned} &2 x+5 y \leq 100 \\ &8 x+5 y \leq 200 \\ &x \geq 0, 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 region are O(0,0), A(25,0), B(50/3,40/3) and C(0,20).

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

Corner Points

Z=x+y

(0,0)

0+0=0

(25,0)

25+0=25

(50/3,40/3)

50/3+50/3=30

(0,20)

0+20=20

 

So the maximum value of Z is 30 at x=\frac{50}{3}, y=\frac{40}{3}

Thus, the maximum distance that the man can travel in one hour 30 km

Hence, the distance travelled by the man at the speed of 225 km/hr is 50/3 km and the distance travelled by him at the speed of the 40km/hour is 40/3 km.

 

Posted by

infoexpert27

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