For the LPP, maximize \( z=x+4 y \) subject to the constraints \( x+2 y \leq 2, x+2 y \geq 8, x, y \geq 0 \)
For the LPP, maximize \( z=x+4 y \) subject to the constraints \( x+2 y \leq 2, x+2 y \geq 8, x, y \geq 0 \)
\( Z_{\max }=4 \)
\( Z_{\max }=8 \)
\( O_{\max }=16 \)
Has no feasible solution
Solution
Given equation, \(z=x+4 y\)
and constraints, \(x+2 y \leq 2 \rightarrow(1)\)
\(\begin{aligned}
&\Rightarrow \frac{x}{2}+\frac{y}{2}=1 \\
&x+2 y \geq 8 \rightarrow(2) \\
&\Rightarrow \frac{x}{8}+\frac{y}{4}=1 \\
&x, y \geq 0 \rightarrow(3)
\end{aligned}\)
Putting \((0,0)\) in Eq. (1), we get \(0 \leq 2\) which is True. Putting \((0,0)\) in Eq. (2), we get \(0 \geq 8\) which is False. Equation (3) implies that the solution is in first quadrant. From the graph, it has no feasible solution.
Putting \((0,0)\) in Eq. (1), we get \(0 \leq 2\) which is True. Putting \((0,0)\) in Eq. (2), we get \(0 \geq 8\) which is False. Equation (3) implies that the solution is in first quadrant. From the graph, it has no feasible solution.