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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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Corner points of the feasible region for an LPP are (0,2)(3,0),(6,0),(6,8) and (0,5).Let $F=4x+6y$ be the objective function,Maximum of F - Minimum of F =

\[(A)\;60\quad(B)\;48\quad(C)\;42\quad(D)\;18\]

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1 Answer

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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 are $(0,2),(3,0),(6,0),(6,8)$ and $(0,5)$
The objective function $F=4x+6y$
At the points $(x,y)$ the objective function subject to $F=4x+6y$
Step 2:
At $(0,2)$ the objective function $F=4x+6y\Rightarrow 4\times 0+6\times 2=12$
At $(3,0)$ the objective function $F=4x+6y\Rightarrow 4\times 3+6\times 0=12$
At $(6,0)$ the objective function $F=4x+6y\Rightarrow 4\times 6+6\times 0=24$
At $(6,8)$ the objective function $F=4x+6y\Rightarrow 4\times 6+6\times 8=72$
At $(0,5)$ the objective function $F=4x+6y\Rightarrow 4\times 0+6\times 5=30$
Step 3:
Maximum of $F$ is 72
Minimum of $F$ is 12
Maximum of $F$ -Minimum of $F =72-12=60$
Hence the correct option is $A$
answered Aug 27, 2013 by sreemathi.v
 

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