Step 1: Rewrite constraints as region descriptions:
The constraint $x+2y \le 120$ means the region lies below or on this line, moving toward the origin side.
The constraint $x+y \ge 60$ means the region lies above or on this line, moving away from the origin side.
The constraint $x-2y \ge 0$, that is $y \le x/2$, means the region lies below or on the line through the origin with slope $1/2$.
Step 2: Test the objective function along the boundary line $x+2y=120$ first:
On this boundary, $Z = 5x+10y = 5(x+2y) = 5(120) = 600$ everywhere on the line, regardless of the exact $x,y$ split.
This tells us immediately that if any part of this line lies in the feasible region, Z equals 600 there, a strong candidate for the maximum since this line is the outer boundary $x+2y\le120$.
Step 3: Find where this line enters and exits the feasible region:
The line $x+2y=120$ must also satisfy $x+y\ge60$ and $x\ge2y$, and $x,y\ge0$.
Setting $y=0$: $x=120$, check $x+y=120\ge60$ true, check $x-2y=120\ge0$ true, so $(120,0)$ is a valid endpoint.
Setting $x=2y$ (meeting the third boundary): $2y+2y=120$ gives $y=30,x=60$, check $x+y=90\ge60$ true, so $(60,30)$ is the other valid endpoint.
Step 4: Confirm no other feasible point gives a higher Z:
Any feasible point satisfies $x+2y\le120$, so $Z=5x+10y=5(x+2y)\le5(120)=600$ always.
This proves 600 is an upper bound for Z over the whole feasible region, not just at corners, confirming the maximum without checking every corner separately.
The bound is achieved exactly on the segment from $(60,30)$ to $(120,0)$, found in Step 3.
Final Answer:
The inequality $Z\le600$ combined with equality on that edge confirms the maximum value.
\[ \boxed{Z_{max} = 600, \text{ attained along the segment joining } (60,30) \text{ and } (120,0)} \]