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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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A company manufactures two types of screws A and B.All the screws have to pass through a threading machine and a slotting machine.A box of type A screws requires 2 minutes on the threading machine and 3minutes on the slotting machine.A box of type B screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine.In a week,each machine is available for 60 hours.On selling these screws,the company gets a profit of Rs100 per box on type A screws and Rs170 per box on type B screws.

$\begin{array}{1 1} (A)\;z=100x+170y \\(B)\;z=170x+100y \\ (C)\;z=3x+4y \\ (D)\;z=4x+3y \end{array} $

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1 Answer

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  • Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
  • If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
The data in the given problem can be summarized as follows :
A box of type A screws requires 2 minutes on the threading machine and 3 minutes on the slotting machine.
A box of type B screws requires 8 minutes on the threading machine and 2 minutes on the slotting machine.
In a week ,each machine is available for 60hours(60$\times$ 60minutes=3600 minutes)
The company gets a profit of Rs100 per box on type A screws and Rs170 per box on type B screws.
Step 2:
Let $x$ be the no of screws required to be on a slotting machine and $y$ be the no of screws required to be on a threading machine.
These screws are brought to fulfill the maximum requirement of $x,y$ and maximize the profit.
Step 3:
The mathematic formulation of the above problem is as follows :
Maximize $z=10x+170y$
Subject to $3x+2y\leq 3600$ and $x+4y\leq 1800,x\geq 0,y\geq 0$
answered Aug 27, 2013 by sreemathi.v
 

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