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  • Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 39

Provide Solution for RD Sharma Class 12 Chapter 29 Linear Programming Exercise 29.4 Question 39

Answers (1)

Answer: The number of book of I type is 12 and II type is 6

Hint:

Let, two types of books be x and y.

Given:

Thickness of the books=6cm and 4 cm

Weight of the books=1 kg and 1 \frac{1}{2} kg each

Shelf is 9 cm and at most can support a weight of 21 kg

Solution:

Let two types of books be x and y respectively.

The required LPP is maximized.

Z=x+y

Subject to the constraints,

6 x+4 y \leq 96\; Or \; 3 x+2 y \leq 48

x+\frac{3}{2} y \leq 21 \text { Or } 2 x+3 y \leq 42

And   x, y \geq 0

On considering the inequalities as equations,

We get,

                                 3 x+2 y=48                                     …(i)

                                 \begin{aligned} &\\ &2 x+3 y=42 \end{aligned}                                     … (ii)

Table for line  3 x+2 y=48  is

 

x

0

16

y

24

0

 

So, it passes through (0, 24) and (16,0)

On putting (0, 0) in 3 x+2 y \leq 48 ,

 We get   0 \leq 48   [Which is true]

Table for  2 x+3 y=42

x

0

21

y

14

0

 

So it passes through (0, 14) and (21, 0)

On putting (0, 0) in 2 x+3 y \leq 42 ,

We get  0 \leq 42  [Which is true]

On solving equation (i) and (ii), we get

                              X = 12 and y = 6

Thus, the intersection point is B (12, 6)

From the graph, OABCD is the feasible region which is bounded. The corner points are O(0,0) A(0,14), B(12,6) and C(16,0)

The values of Z at corner points are as follows.

Corner Points

Z=x+y

O(0,0)

Z=0+0=0

A(0,14)

Z=0+14=14

B(12,6)

Z=12+6=18

C(16,0)

Z=16+0=16

 

From the table, the maximum value of Z is 18 at B(12,6)

Hence, the maximum number of books is 18 and number of books of I type is 12 and books of II type is 6.

 

 

Posted by

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