# The feasible solution for a LPP is shown in fig.12.12.Let $Z=3x-4y$ be the objective function.Minimum of $Z$ occurs at

> $(A)(0,0) \quad (B)\; (0,8)\quad (C)\;(5,0) \quad (d)\;(4,10)$

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 objective function is $Z=3x-4y$
The corner points are $(0,0),(5,0),(6,5),(4,10),(0,8)$
For the points $(x,y)$ the objective function subject to $Z=3x-4y$
Step 2:
At $(0,0)$ the objective function $Z=3x-4y\Rightarrow Z=0$
At $(5,0)$ the objective function $Z=3x-4y\Rightarrow 3\times 5-4\times 0=15$
At $(6,5)$ the objective function $Z=3x-4y\Rightarrow 3\times 6-4\times 5=-2$
At $(6,8)$ the objective function $Z=3x-4y\Rightarrow 3\times 6-4\times 8=-14$
At $(4,10)$ the objective function $Z=3x-4y\Rightarrow 3\times 4-4\times 10=-28$
At $(0,8)$ the objective function $Z=3x-4y\Rightarrow 3\times 0-4\times 8=-32$
Step 3:
The minimum $Z$ occurs at $(0,8)$ which is $-32$
Hence the correct option is $B$