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  • Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 55

Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 55

Answers (1)

Answer: Maximum Z=100x+300y is the required LPP.

Hint:

 Use property of LPP

Given:

The maximum number of hours available per day is16, If the profit on a necklace is Rs.100 and the bracelet is Rs.300

Solution:

Let the number of necklaces manufacturer be x and the number of bracelets manufacture be y.

Since the total number of item are at most 24.

       x+y \leq 24                                    … (i)

Bracelets take 1 hour to manufacture and necklaces take half an hour to manufacture.

X item take x hour to manufacture and y item take y/2 hours to manufacture and maximum time available is 16 hours

Therefore,

      \frac{x}{2}+y \leq 16                                    … (ii)

The profit on one necklace is Rs.100 and the profit on one bracelet is Rs.300.

Let the profit be Z. Now we wish to maximize the profit.

So,

     Maximize Z=100x + 300y          … (iii)

So,

       \begin{aligned} &x+y \leq 24 \\ &\frac{x}{2}+y \leq 16 \end{aligned}

Maximize Z = 100x + 300y is required LPP

X+y=24

 

Corner points

Max Z

(0,16)

1800

(24,0)

2400

(16,18)

7000

 

           The maximum value of Z = 7000 at (16,18)

Posted by

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