The solution of the linear programming proble | vedclass

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(B) To find the maximum value of $Z = 3x_{1} + 5x_{2}$,we evaluate the objective function at the corner points of the feasible region defined by the c

Vedclass · Accepted Answer

$x_{1} = 2, x_{2} = 6, Z = 36$

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(B) To find the maximum value of $Z = 3x_{1} + 5x_{2}$,we evaluate the objective function at the corner points of the feasible region defined by the constraints:$1$. $3x_{1} + 2x_{2} \leq 18$$2$. $x_{1} \leq 4$$3$. $x_{2} \leq 6$$4$. $x_{1} \geq 0, x_{2} \geq 0$The corner points are found by the intersection of these lines:- Intersection of $x_{1} = 0$ and $x_{2} = 0$ is $(0, 0)$. $Z = 0$.- Intersection of $x_{1} = 4$ and $x_{2} = 0$ is $(4, 0)$. $Z = 3(4) + 5(0) = 12$.- Intersection of $x_{1} = 4$ and $3x_{1} + 2x_{2} = 18$ gives $3(4) + 2x_{2} = 18 \Rightarrow 2x_{2} = 6 \Rightarrow x_{2} = 3$. Point is $(4, 3)$. $Z = 3(4) + 5(3) = 12 + 15 = 27$.- Intersection of $3x_{1} + 2x_{2} = 18$ and $x_{2} = 6$ gives $3x_{1} + 2(6) = 18 \Rightarrow 3x_{1} = 6 \Rightarrow x_{1} = 2$. Point is $(2, 6)$. $Z = 3(2) + 5(6) = 6 + 30 = 36$.- Intersection of $x_{1} = 0$ and $x_{2} = 6$ is $(0, 6)$. $Z = 3(0) + 5(6) = 30$.Comparing the values of $Z$ at all corner points,the maximum value is $36$ at $(2, 6)$.

The solution of the linear programming problem,maximize $Z = 3x_{1} + 5x_{2}$ subject to $3x_{1} + 2x_{2} \leq 18$,$x_{1} \leq 4$,$x_{2} \leq 6$,$x_{1} \geq 0$,$x_{2} \geq 0$ is:

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