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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.Solve the linear programming problem and determine the maximum profit to the manufacturer.

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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 given problem can be written as follows :
Type A screws :Threading machine requires 2 min,slotting machine requires 3 min,cost per box =Rs.100
Type B screws :Threading machine requires 8 min,slotting machine requires 2 min,cost per box =Rs.170
The mathematical formulation for the above problem is as follows :
Let $x$ be the no of screws of type A
Let $y$ be the no of screws of type B
Time taken for the screws of type A and type B to be allowed in threading machine is 2min and 8 min respectively.
Time taken for the screws of type A and type B to be allowed in slotting machine is 3min and 2 min respectively.
The time available in one week is 60 hour (i.e.)3600min
The cost per box for type A screws is Rs 100 and type B box is Rs.170
We have the maximize the profit
Hence the objective function is $Z=100x+170y$ subjected to constraints.
$2x+8y\leq 3600,3x+2y\leq 3600$ and $x,y\geq 0$
Step 2:
Let us draw the graph for the lines $AB :2x+8y=3600$ and $CD :3x+2y=3600$
Consider the line $AB :2x+8y=3600$
Put $x=0,y=0$ then $0\leq 3600$ is true.
Clearly the region $2x+8y\leq 3600$ lies below the line AB.
Consider the line $CD :3x+2y=3600$
Put $x=0,y=0$ then $0\leq 3600$ is true.
Hence the region $3x+2y\leq 3600$ lies below the line CD.
The origin O(0,0) is also included in the feasible region.
The feasible region OCPB is the shaded portion shown in the fig.
The point of intersection of the two lines is $P(1080,180)$
Hence the corner points of the feasible region OCPB are $O(0,0),C(0,450),P(1080,180),B(1200,0)$
Step 3:
To find the value of the objective function $Z=100x+170y$ as follows :
At the points $(x,y)$ the value of the objective function subjected to $Z=100x+170y$
At $O(0,0)$ the value of the objective function $Z=0$
At $C(0,450)$ the value of the objective function $Z=100(0)+170(450)=76500$
At $P(1080,180)$ the value of the objective function $Z=100(1080)+170(180)=138600$
At $B(1200,0)$ the value of the objective function $Z=100(1200)+0=120000$
Hence the maximum value of $Z$ is at $P(1080,180)$ .
The maximum profit is Rs.138600$
answered Aug 30, 2013 by sreemathi.v
 

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