Question

The point which does not belong to the feasible region of the LPP:Minimize: $\mathrm{Z}=60\mathrm{x}+10\mathrm{y}$subject to $3\mathrm{x}+\mathrm{y}\ge 18$$2x+2y\ge 12$$x+2y\ge 10$$\mathrm{x},\mathrm{y}\ge 0$ is:

• (A) $(0,8)$
• (B) $(4,2)$
• (C) $(6,2)$
• (D) $(10,0)$

Answer 1

The Correct Option is B

Explanation:
Given:Minimize: $\mathrm{Z}=60\mathrm{x}+10\mathrm{y}$subject to $3\mathrm{x}+\mathrm{y}\ge 18$$2x+2y\ge 12$$x+2y\ge 10$$\mathrm{x},\mathrm{y}\ge 0$We test whether the inequalitiies are satisfied or not $(0,8),3(0)+8\ge 88\ge 8$ is true.$2(0)+2(8)=16\ge 12$ is true. $0+2(8)=16\ge 10$ is true.$\therefore (0,8)$ is in the feasible region.$(4,2),3(4)+2=14\ge 8$$2(4)+2(2)=16\ge 12$$4+2(2)=8\ge 10$ is not true$\therefore (4,2)$ is not a point in the feasible regionHence, the correct option is (B).