The maximum value of Z=5x+4y,subject to the c | vedclass

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(A) We are given the objective function $Z = 5x + 4y$ subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,and $y \geq 0$.First,

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$14$

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(A) We are given the objective function $Z = 5x + 4y$ subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,and $y \geq 0$.First,we identify the feasible region by plotting the lines $y = 2x$,$x = 2y$,and $x+y = 3$.$1$. The intersection of $y = 2x$ and $x+y = 3$: Substituting $y=2x$ into $x+y=3$ gives $x+2x=3$,so $3x=3$,which means $x=1$. Then $y=2(1)=2$. So,the point is $A(1, 2)$.$2$. The intersection of $x = 2y$ and $x+y = 3$: Substituting $x=2y$ into $x+y=3$ gives $2y+y=3$,so $3y=3$,which means $y=1$. Then $x=2(1)=2$. So,the point is $B(2, 1)$.$3$. The origin $O(0, 0)$ is also a corner point.The feasible region is the triangle $OAB$. We evaluate $Z$ at the corner points:Corner Point$Z = 5x + 4y$$O(0, 0)$$5(0) + 4(0) = 0$$A(1, 2)$$5(1) + 4(2) = 5 + 8 = 13$$B(2, 1)$$5(2) + 4(1) = 10 + 4 = 14$Thus,the maximum value of $Z$ is $14$ at point $B(2, 1)$.

The maximum value of $Z=5x+4y$,subject to the constraints $y \leq 2x$,$x \leq 2y$,$x+y \leq 3$,$x \geq 0$,$y \geq 0$ is:

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