The objective function Z = 4 x_1 + 5 x_2,subj | vedclass

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(A) The objective function is $Z = 4 x_1 + 5 x_2$.The constraints are:$1) 2 x_1 + x_2 \geq 7$$2) 2 x_1 + 3 x_2 \leq 15$$3) x_2 \leq 3$$4) x_1, x_2 \ge

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On $x_1$-axis

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(A) The objective function is $Z = 4 x_1 + 5 x_2$.The constraints are:$1) 2 x_1 + x_2 \geq 7$$2) 2 x_1 + 3 x_2 \leq 15$$3) x_2 \leq 3$$4) x_1, x_2 \geq 0$To find the feasible region,we determine the intersection points of the boundary lines:For $2 x_1 + x_2 = 7$,the intercepts are $(3.5, 0)$ and $(0, 7)$.For $2 x_1 + 3 x_2 = 15$,the intercepts are $(7.5, 0)$ and $(0, 5)$.The line $x_2 = 3$ is a horizontal line.Solving for intersection points of the feasible region:- Intersection of $2 x_1 + x_2 = 7$ and $x_2 = 3$: $2 x_1 + 3 = 7 \implies 2 x_1 = 4 \implies x_1 = 2$. Point: $(2, 3)$.- Intersection of $2 x_1 + 3 x_2 = 15$ and $x_2 = 3$: $2 x_1 + 9 = 15 \implies 2 x_1 = 6 \implies x_1 = 3$. Point: $(3, 3)$.- The $x_1$-intercepts are $(3.5, 0)$ and $(7.5, 0)$.The corner points of the feasible region are $(3.5, 0), (7.5, 0), (3, 3), (2, 3)$.Evaluating $Z = 4 x_1 + 5 x_2$ at these points:Corner Point$Z = 4 x_1 + 5 x_2$$(3.5, 0)$$4(3.5) + 5(0) = 14$$(7.5, 0)$$4(7.5) + 5(0) = 30$$(3, 3)$$4(3) + 5(3) = 12 + 15 = 27$$(2, 3)$$4(2) + 5(3) = 8 + 15 = 23$The minimum value of $Z$ is $14$,which occurs at the point $(3.5, 0)$. Since the $x_2$-coordinate is $0$,this point lies on the $x_1$-axis.

The objective function $Z = 4 x_1 + 5 x_2$,subject to $2 x_1 + x_2 \geq 7$,$2 x_1 + 3 x_2 \leq 15$,$x_2 \leq 3$,$x_1, x_2 \geq 0$ has minimum value at the point

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