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Questions  >>  CBSE XII  >>  Math  >>  Linear Programming

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 . The minimum value of F occurs at\[\begin{array}{1 1}(A)\;(0,2)\;only \\ (B)\;(3,0)\;only\\(C)\;the\;mid\;point\;of\;the\;line\;segment\;joining\;the\;points\;(0,2) \;and\;(3,0)\;only\\(D)\;any\;point\;on\;the\;line\;segment\;joining\;the\;points\;(0,2) \;and\;(3,0)\end{array}\]

1 Answer

  • 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 $(0,2),(3,0),(6,0)$ 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:
The minimum value of the objective function is 12 and this occurs at both the point $(0,2)$ and $(3,0)$
Hence the minimum value of $F$ occurs at any point on the line segment joining the points $(0,2)$ and $(3,0)$
The correct option is $D$
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