The L.P.P. to maximize z=x+y,subject to x+y l | vedclass

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(B) To find the solution,we first identify the feasible region defined by the constraints:$1$. $x+y \leq 30$$2$. $x \leq 15$$3$. $y \leq 20$$4$. $x+y

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a unique solution.

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(B) To find the solution,we first identify the feasible region defined by the constraints:$1$. $x+y \leq 30$$2$. $x \leq 15$$3$. $y \leq 20$$4$. $x+y \geq 15$$5$. $x, y \geq 0$The vertices of the feasible region are determined by the intersection of these lines:- Intersection of $x=15$ and $y=20$ is $E(15, 20)$.- Intersection of $x=0$ and $y=20$ is $D(0, 20)$.- Intersection of $x=0$ and $x+y=15$ is $F(0, 15)$.- Intersection of $x=15$ and $x+y=15$ is $C(15, 0)$.The feasible region is the quadrilateral $CDEF$.We evaluate the objective function $z=x+y$ at these vertices:- At $C(15, 0): z = 15+0 = 15$- At $D(0, 20): z = 0+20 = 20$- At $E(15, 20): z = 15+20 = 35$- At $F(0, 15): z = 0+15 = 15$The maximum value of $z$ is $35$,which occurs at the unique vertex $E(15, 20)$. Therefore,the $L$.$P$.$P$. has a unique solution.

The $L$.$P$.$P$. to maximize $z=x+y$,subject to $x+y \leq 30, x \leq 15, y \leq 20, x+y \geq 15$,and $x, y \geq 0$ has

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