Step 1: Understand the goal.
We minimise $z=4x+6y$ over the region defined by $x+2y\ge80$, $3x+y\ge75$, and $x,y\ge0$. The minimum of a linear function over such a region sits at a corner point.
Step 2: List the boundary lines.
Treat the constraints as equalities: $L_1:x+2y=80$ and $L_2:3x+y=75$.
Step 3: Find the corner on the axes.
$L_1$ meets the y-axis at $(0,40)$ and the x-axis at $(80,0)$; $L_2$ meets the axes at $(0,75)$ and $(25,0)$. Because both inequalities are $\ge$, the feasible region is the unbounded part away from the origin, whose corners are $(80,0)$, $(0,75)$, and the meeting point of the two lines.
Step 4: Solve for the intersection.
From $L_2$, $y=75-3x$. Put into $L_1$: $x+2(75-3x)=80$, so $x+150-6x=80$, giving $-5x=-70$ and $x=14$. Then $y=75-42=33$. The corner is $(14,33)$.
Step 5: Evaluate $z$ at each corner.
At $(80,0)$: $z=320$. At $(0,75)$: $z=450$. At $(14,33)$: $z=4(14)+6(33)=56+198=254$.
Step 6: Pick the smallest.
The least value among $320$, $450$, and $254$ is $254$, occurring at $(14,33)$, matching option (4).
\[ \boxed{z_{\min}=254} \]