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  • Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 5

Explain solution RD Sharma class 12 Chapter 29 Linear Programming exercise 29.4 question 5

Answers (1)

Answer:

Required number belt A is 200, while B is 600, maximum profit Rs1300.

Hint:

Let required number of belt A and B be x and y.

Given:

Profit on belt A and B be Rs2 and Rs1.50

Solution:

Let z be total profit

z =2x+1.5y

Where x and y be required number of belt A and belt B.

Since each belt of type A required twice as much time as belt B. let each belt B require to make, so A requires 2 hours but total time available is equal to production 1000 belt B that is 1000 hours, so

2x+y\leq 1000  (first constraint)

Given supply of lather only for 800 belts per day both A and B combined, so

x+y\leq 800 (second constraint)

Buckets available for A is only 400 and for B only 700, so

x\leq 400 (third constraint)

y\leq 700  (forth constraint)

Hence the required LPP

z =2x+1.5y

Subject to constraints

2x+y\leq 1000\\       ....(i)

x+y\leq 800\\        ....(ii)

x\leq 400             .....(iii)

y\leq 700          ...(iv)

x,y\geq 0,number of belt  cannot less than zero

The feasible region determined by the system of constraints

Obtain point Q(200,600) by solving (i) and (ii) and P(400,200) by solving (i) and (iii) and R(100,700) by solving (ii) and (iv)

The shading region is  OA_{3}PQRB_{2}

Corner Points

Value of  z=2x+1.5y

O\left ( 0,0 \right )

0

A_{3}\left (400,0 \right )

600

P(400,200)

1100

Q\left (200,600 \right )

1300

R\left (0,700 \right )

1050

Maximum profit = 1300, required number belt A = 200, belt B = 600

Posted by

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