The maximum value of z=6x+8y subject to the c | vedclass

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(B) The objective function is $z=6x+8y$. The constraints are $x-y \geq 0$,$x+3y \leq 12$,$x \geq 0$,and $y \geq 0$. To find the feasible region,we plo

Vedclass · Accepted Answer

$42$

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(B) The objective function is $z=6x+8y$. The constraints are $x-y \geq 0$,$x+3y \leq 12$,$x \geq 0$,and $y \geq 0$. To find the feasible region,we plot the lines $x-y=0$ and $x+3y=12$. The intersection point of $x-y=0$ and $x+3y=12$ is found by substituting $x=y$ into $x+3y=12$,which gives $4y=12$,so $y=3$ and $x=3$. Thus,the intersection point is $B(3, 3)$. The feasible region is a triangle with vertices $O(0, 0)$,$A(0, 4)$ (from $x+3y=12$ when $x=0$),and $B(3, 3)$. Evaluating $z=6x+8y$ at these corner points: At $O(0, 0)$: $z = 6(0) + 8(0) = 0$. At $A(0, 4)$: $z = 6(0) + 8(4) = 32$. At $B(3, 3)$: $z = 6(3) + 8(3) = 18 + 24 = 42$. The maximum value is $42$.

The maximum value of $z=6x+8y$ subject to the constraints $x-y \geq 0$,$x+3y \leq 12$,$x \geq 0$,$y \geq 0$ is:

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