The LPP to maximize z = 2x + 5y subject to th | vedclass

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(A) Step $1$: Analyze the constraints. The constraints are $x + 3y \leq 6$ and $2x + 6y \leq 18$. Note that $2x + 6y \leq 18$ simplifies to $x + 3y \l

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(A) Step $1$: Analyze the constraints. The constraints are $x + 3y \leq 6$ and $2x + 6y \leq 18$. Note that $2x + 6y \leq 18$ simplifies to $x + 3y \leq 9$. Step $2$: Since $x + 3y \leq 6$ is a stricter condition than $x + 3y \leq 9$, the feasible region is determined solely by $x + 3y \leq 6$ along with $x \geq 0$ and $y \geq 0$. Step $3$: The vertices of the feasible region are $(0, 0)$, $(6, 0)$, and $(0, 2)$. Step $4$: Evaluate $z = 2x + 5y$ at these vertices: At $(0, 0)$, $z = 2(0) + 5(0) = 0$. At $(6, 0)$, $z = 2(6) + 5(0) = 12$. At $(0, 2)$, $z = 2(0) + 5(2) = 10$. Step $5$: The maximum value is $12$ at the point $(6, 0)$. Since there is a single point where the maximum is attained, the solution is unique.

The $LPP$ to maximize $z = 2x + 5y$ subject to the constraints $x + 3y \leq 6$, $2x + 6y \leq 18$, $x \geq 0$, $y \geq 0$ has:

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