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.MathematicsRS AgarwalStandard XII

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