Maximum value of $\mathrm{z}=3 x+4 \mathrm{y}$ subject to $x-\mathrm{y} \leqslant-1, \quad-x+\mathrm{y} \leqslant 0, \quad x, \mathrm{y} \geqslant 0$ is given by

To determine the maximum value of $z = 3x + 4y$ under the constraints, analyze the feasible region.

The first constraint implies $y \geq x + 1$, and the second implies $y \leq x$. For any $(x,y)$ to satisfy both, $x + 1 \leq y \leq x$ must hold, leading to $x + 1 \leq x$, which simplifies to $1 \leq 0$. This is a contradiction, so no such point exists.

With no feasible region, the objective function has no maximum value.

The answer is $\boxed{D}$.