The point at which the maximum value of $x + y$ subject to constraints $x + 2y \leq 70$ and $2x + y \leq 95, x \geq 0, y \geq 0$ is.

The objective function to be maximized is $Z = x + y$.

The given constraints are:
$x + 2y \leq 70$
$2x + y \leq 95$
$x \geq 0, y \geq 0$

The corner points of the feasible region are determined by the intersections of the boundary lines $x + 2y = 70$, $2x + y = 95$, and the coordinate axes.

Solving $x + 2y = 70$ and $2x + y = 95$:
Multiplying the second equation by $2$ gives $4x + 2y = 190$.
Subtracting the first equation from this gives $3x = 120 \Rightarrow x = 40$.
Substituting $x = 40$ into $x + 2y = 70$ gives $40 + 2y = 70 \Rightarrow 2y = 30 \Rightarrow y = 15$.
The intersection point is $(40, 15)$.

The corner points of the bounded feasible region are $(0, 0)$, $(47.5, 0)$, $(40, 15)$, and $(0, 35)$.

Evaluating $Z = x + y$ at each corner point:
At $(0, 0)$, $Z = 0 + 0 = 0$
At $(47.5, 0)$, $Z = 47.5 + 0 = 47.5$
At $(40, 15)$, $Z = 40 + 15 = 55$
At $(0, 35)$, $Z = 0 + 35 = 35$

The maximum value of $Z$ is $55$, which occurs at the point $(40, 15)$.

Answer: $(40, 15)$