The linear programming problem (LPP) to maxim | vedclass

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(A) Step $1$: Identify the feasible region defined by the constraints.Constraint $1$: $x + 3y \leq 6$. The boundary line passes through $(6, 0)$ and $

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$A$ unique optimal solution

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(A) Step $1$: Identify the feasible region defined by the constraints.Constraint $1$: $x + 3y \leq 6$. The boundary line passes through $(6, 0)$ and $(0, 2)$.Constraint $2$: $2x + 6y \leq 18$ simplifies to $x + 3y \leq 9$. The boundary line passes through $(9, 0)$ and $(0, 3)$.Step $2$: Since $x + 3y \leq 6$ is a stricter constraint than $x + 3y \leq 9$, the feasible region is determined by $x + 3y \leq 6$ in the first quadrant.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 the maximum occurs at a single vertex, the solution is unique.

Column	Value
$A$. Maximum of $z$	$300$
$B$. Constant value	$325$

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

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Column Value

$A$. Maximum of $z$ $300$

$B$. Constant value $325$

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