Maximum value of z = 3x + 4y subject to the c | vedclass

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(D) The given constraints are:$1) x - y \leqslant -1 \implies y \geqslant x + 1$$2) -x + y \leqslant 0 \implies y \leqslant x$$3) x, y \geqslant 0$Fro

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Does not exist

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(D) The given constraints are:$1) x - y \leqslant -1 \implies y \geqslant x + 1$$2) -x + y \leqslant 0 \implies y \leqslant x$$3) x, y \geqslant 0$From constraint $(1)$,$y \geqslant x + 1$. Since $x \geqslant 0$,the minimum value of $y$ is $1$. From constraint $(2)$,$y \leqslant x$. Combining these,we get $x + 1 \leqslant y \leqslant x$. This implies $x + 1 \leqslant x$,which simplifies to $1 \leqslant 0$. This is a contradiction,meaning there is no point $(x, y)$ that satisfies all the given constraints simultaneously. Therefore,the feasible region is empty,and the maximum value of $z$ does not exist.

Maximum value of $z = 3x + 4y$ subject to the constraints $x - y \leqslant -1$,$-x + y \leqslant 0$,and $x, y \geqslant 0$ is:

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