Step 1: List candidate vertices from the constraint boundaries and axes:
$(0,0)$, the two axis-intercepts of each line, and their mutual intersection are the only places a linear objective can peak.
Step 2: Discard intercepts that violate the other constraint:
$(0,5)$ fails $2x+5y\le10$ (gives 25); $(5,0)$ fails $5x+3y\le15$ (gives 25). Only $(3,0)$ and $(0,2)$ survive as feasible intercepts.
Step 3: Solve the 2x2 system for the intersection point:
$5x+3y=15$ and $2x+5y=10$. Multiply first by 5 and second by 3: $25x+15y=75$, $6x+15y=30$. Subtract: $19x=45\Rightarrow x=45/19$, then $y=20/19$.
Step 4: Compare Z at all four feasible vertices:
$(0,0)\to0$; $(3,0)\to30$; $(0,2)\to6$; $(45/19,20/19)\to510/19\approx26.8$. The largest is at $(3,0)$.
Final Answer:
\[ \boxed{Z_{max}=30\text{ at }(3,0)} \]