Maximum value of Z=100 x+70 y subject to 2 x | vedclass

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(A) The constraints are $2x \geq 4$ (or $x \geq 2$),$y \leq 3$,$x+y \leq 8$,and $x, y \geq 0$.The feasible region is bounded by the lines $x=2, y=3, x

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(A) The constraints are $2x \geq 4$ (or $x \geq 2$),$y \leq 3$,$x+y \leq 8$,and $x, y \geq 0$.The feasible region is bounded by the lines $x=2, y=3, x+y=8$ and the $x$-axis $(y=0)$.The corner points of the feasible region are determined by the intersection of these lines:$A(2, 0)$ (intersection of $x=2$ and $y=0$)$B(8, 0)$ (intersection of $x+y=8$ and $y=0$)$C(5, 3)$ (intersection of $x+y=8$ and $y=3$)$D(2, 3)$ (intersection of $x=2$ and $y=3$)Now,we evaluate $Z=100x+70y$ at each corner point:At $A(2, 0): Z = 100(2) + 70(0) = 200$At $B(8, 0): Z = 100(8) + 70(0) = 800$At $C(5, 3): Z = 100(5) + 70(3) = 500 + 210 = 710$At $D(2, 3): Z = 100(2) + 70(3) = 200 + 210 = 410$The maximum value of $Z$ is $800$ at point $B(8, 0)$.

Maximum value of $Z=100 x+70 y$ subject to $2 x \geq 4, y \leq 3, x+y \leq 8, x, y \geq 0$ is

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