z = 30x+20y, x+y ≤ 8, x + 2y ≥ 4, 6x + 4y ≥ 12, x≥0, y ≥ 0 has

Since, $x+y=8$       ....(i)

This line meets axes at (8,0) and (0,8) respectively.

$x+2y=4$        .....(ii)

$⇒\frac{x}{4}+\frac{y}{2}=1$

This line meet axes at (4,0) and (0,2).

And $6x+4y=12$        ......(iii)

$⇒\frac{x}{2}+\frac{y}{3}=1$

This line meets axes at (2,0) and (0,3)

The point of intersection of equations (ii) and (iii) is $F(1,\frac{3}{2})$

Now, at A(4,0), $z=30×4=120$

             B(8,0), $z=30×8=240$

             C(0,8), $z=20×8=160$

             D(0,3), $z=20×3=60$

and $F(1,\frac{3}{2}),z=30×1+20×\frac{3}{2}=60$

It is clear that z is minimum 60 at points D(0,3) and $F(1,\frac{3}{2})$

Since minimum value is at two points

there is infinitely many solutions.