The L.P.P. , minimize $z=30 x+20 y, \quad x+y \leqslant 8, \quad x+2 y \geqslant 4, \quad 6 x+4 y \geqslant 12$, $x \geqslant 0, y \geqslant 0$ has

Linear Programming Problem: Minimize $z = 30x + 20y$ subject to the constraints:

$x + y \leq 8$
$x + 2y \geq 4$
$3x + 2y \geq 6$
$x \geq 0$, $y \geq 0$

The feasible region lies in the first quadrant, bounded by the lines $x + y = 8$, $x + 2y = 4$, and $3x + 2y = 6$. Intersections of these constraints yield the vertices:

From $x = 0$ and $3x + 2y = 6$, $y = 3$: $(0, 3)$.
From $x = 0$ and $x + y = 8$, $y = 8$: $(0, 8)$.
From $y = 0$ and $x + 2y = 4$, $x = 4$: $(4, 0)$.
From $y = 0$ and $x + y = 8$, $x = 8$: $(8, 0)$.
From $x + 2y = 4$ and $3x + 2y = 6$, solving gives $x = 1$, $y = \frac{3}{2}$: $(1, \frac{3}{2})$.

Evaluating the objective function:

At $(0, 3)$: $z = 30(0) + 20(3) = 60$
At $(0, 8)$: $z = 30(0) + 20(8) = 160$
At $(4, 0)$: $z = 30(4) + 20(0) = 120$
At $(8, 0)$: $z = 30(8) + 20(0) = 240$
At $(1, \frac{3}{2})$: $z = 30(1) + 20(\frac{3}{2}) = 60$

The minimum value of $z$ is 60, occurring at both $(0, 3)$ and $(1, \frac{3}{2})$. Since these are adjacent vertices with identical objective values, the line segment between them (part of $3x + 2y = 6$) yields infinitely many solutions achieving the minimum.

Thus, the problem has infinitely many optimal solutions.