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Question

A manufacturing company makes two types of teaching aids $$A$$ and $$B$$ of Mathematics for Class XII. Each type of $$A$$ requires $$9$$ labour hours of fabricating and $$1$$ labour hour for finishing. Each type of $$B$$ requires $$12$$ labour hours for fabricating and finishing, the maximum labour hours available per week are $$180$$ and $$30$$ respectively. The company makes a profit of $$Rs. 80$$ on each piece of type $$A$$ and $$Rs. 120$$ on each piece of type $$B$$. How many pieces of  type $$B$$ should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?


Solution

 Fabricating Hours Finishing Hours 
 A$$9$$ $$1$$ 
 $$B$$$$12$$ $$3$$ 

Let pieces of type A manufactured per week $$= x$$
Let pieces of type B manufactured per week $$= y$$
Companies profit function which is to be maximized: Z $$= 80x + 120y$$ as shown in the tabular column:
Constraints: Maximum number of fabricating hours $$= 180$$
Therefore, $$9x+12y\leq 180\Rightarrow 3x+4y\leq 60$$
Where $$9x$$ is the fabricating hours spent by type A teacher aids, and $$12y$$ hours spent on type B and maximum number of finishing hours $$= 30.  $$

$$x+3y\leq 30$$
Where x is the number of hours spent on finishing aid A while 3y on aid B.
So, the LPP becomes:
$$Z(maximise)=80x+120y$$
Subject to $$3x+4y\leq 60$$
$$x+3y\leq30$$
$$x\geq 0$$
$$y\geq 0$$
Solving it graphically:
Z$$=80x+120y$$ at $$(0, 15) = 1800$$
Z $$= 1200$$ at $$(0, 10)$$
Z $$=1600$$ at $$(20, 0)$$
Z $$= 960 + 720$$ at $$(12, 6) = 1680$$
Maximum profit is at $$(0, 15).$$
Therefore, Teacher aid A $$= 0$$
Teacher aid B $$= 15$$ should be made.

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Mathematics
RS Agarwal
Standard XII

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