Linear Programming Test

Result

Linear Programming Test - 14

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Question 1
1 / -0

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

A
\((0,8)\)

B
\((4,2)\)

C
\((6,2)\)

D
\((10,0)\)

Solution

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