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Please Solve RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.5 Question 1 Maths Textbook Solution.

Answers (1)

Answer:

                Minimum transportation cost = 150

                From A: 10,50,40 \text { quintals to } D, E, F

                From  B: 50,0,0 \text { quintals to } D, E, F

Hint:

Draw the graph and find the value of z

Given:

 

Transportation

 

Cost per quintal

From

A

B

To

D

E

F

6.00

3.00

2.50

4.00

2.00

3.00

 

Let the supply of wheat is x quintal from  to A\: to\: D and Y quintal from A to E .Then wheat supply will be (100-x-y) quintal from A to F.

Similarly, (60-x),(50-y),(x+y-60) quintals of wheat will be supplied from B\: to\: D, E, F respectively.

                      

Where, N represents the Quintals

Now minimum transportation cost,

                \begin{aligned} &z=6 x+3 y+2.50(100-x-y)+4(600-x)+2(50-y)+3(x+y-60) \\ & \end{aligned}

                z=2.50 x+1.50 y+410

And constant  x \geq 0, y \geq 0

                \begin{aligned} &100-x-y \geq 0 \Rightarrow x+y \leq 100 \\ & \end{aligned}

                60+x \geq 0 \Rightarrow x \leq 60 \\

                50-y \geq 0 \Rightarrow y \leq 50 \\

                x+y-60 \geq 0 \Rightarrow x+y \geq 60

First, we draw the graph of the lines,

                x+y=100, x=60, y=50, x+y=60

              

Now, we find the feasible region by constants

                  x+y \leq 100, x \leq 60, y \leq 50, x+y \geq 60, x \geq 0, y \geq 0

And shade it. Its vertices are

                A(10,50), B(60,0), C(60,40), D(50,50)  at which we find  z

Vertices

coordinates

z=2.50 x+1.50 y+410

A

B

C

D

\left ( 10,50 \right )

\left ( 60,0 \right )

\left ( 60,40 \right )

\left ( 50,50 \right )

510-minimum

560

620

610

 

Therefore, minimum transportation cost Rs.510. For this 10,50,40 quintals will supply from A \: to \: D, E, F \text { and } 50,0,0 quintals will supply B \: to\: D, E, F.

 

Posted by

infoexpert27

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