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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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The feasible region for an LPP is shown in the figure below ( fig 12.13 in textbook ).Let $F=3x-4y$ be the objective function.Minimum value of $F$ is

\[(A)\;0 \quad (B)\; -16\quad (C)\;12 \quad (D)\;does\;not\;exist\]

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Toolbox:
  • 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 corner points of the feasible region are $(6,0)$ and $(12,6)$
The given objective function $F=3x-4y$ has minimum value.
For the points $(x,y)$ the objective function subject to $F=3x-4y$
Step 2:
At $(6,0)$ the objective function $F=3x-4y\Rightarrow 3\times 6-4\times 0=18$
At $(12,6)$ the objective function $F=3x-4y\Rightarrow 3\times 12-4\times 6=12$
Hence the minimum value of $F$ is 12
The correct option is $C$
answered Aug 27, 2013 by sreemathi.v
 

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