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Home » A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of each on type and each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of \pm 50 each on type A and \pm 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?
CBSE | Class 12 | Exercise 33B | Linear Programming | Maths | RS Aggarwal
A company manufacture two types of toys A and B. type A requires 5 minutes each for cutting and 10 minutes for each assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. He earns a profit of \pm 50 each on type A and \pm 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?

Let the company manufacture x and y numbers of toys A and B.
\therefore According to the question,
5 x+8 y \leq 180,10 x+8 y \leq 240, x \geq 0, y \geq 0
Maximize Z=50 x+60 y
The feasible region determined 5 x+8 y \leq 180,10 x+8 y \leq 240, x \geq 0, y \geq 0 is given by

The corner points of feasible region are A(0,0), B(0,22.5), C(12,15), D(24,0) . The value of Z at corner point is

The maximum value of Z is 1500 and occurs at point (12,15).
The company should manufacture 12 A toys and 15 B toys to earn profit of rupees 1500 .

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