$\text { The maximum value of } Z=3 x+4 y \text { for the given constraints } x+2 y \leq 76,2 x+y \leq 104, x \geq 0, y \geq 0 \text { is }$

The constraints are $x + 2y \leq 76$, $2x + y \leq 104$, $x \geq 0$, and $y \geq 0$.

The corner points of the feasible region are found by solving the intersection of the boundary lines and the axes.

1. Intersection of $x + 2y = 76$ and $2x + y = 104$:
Multiply the first equation by 2: $2x + 4y = 152$.
Subtract the second equation: $(2x + 4y) - (2x + y) = 152 - 104 \Rightarrow 3y = 48 \Rightarrow y = 16$.
Substitute $y = 16$ into $x + 2(16) = 76 \Rightarrow x + 32 = 76 \Rightarrow x = 44$.
Corner point: $(44, 16)$.

2. Intersection of $x + 2y = 76$ with the y-axis ($x=0$):
$2y = 76 \Rightarrow y = 38$.
Corner point: $(0, 38)$.

3. Intersection of $2x + y = 104$ with the x-axis ($y=0$):
$2x = 104 \Rightarrow x = 52$.
Corner point: $(52, 0)$.

4. Origin: $(0, 0)$.

Evaluate $Z = 3x + 4y$ at each corner point:
At $(0, 0)$: $Z = 3(0) + 4(0) = 0$.
At $(0, 38)$: $Z = 3(0) + 4(38) = 152$.
At $(52, 0)$: $Z = 3(52) + 4(0) = 156$.
At $(44, 16)$: $Z = 3(44) + 4(16) = 132 + 64 = 196$.

The maximum value is 196.

Answer: 196