Step 1: Use a simple rewrite first.
Write \(Z=38x+19y=19(2x+y)\). So we need to maximise \(2x+y\) and multiply by 19.
Step 2: Bound 2x + y using the constraints.
We look for numbers \(\lambda,\mu\ge0\) with \(\lambda(3x+5y)+\mu(5x+2y)=2x+y\). This needs \(3\lambda+5\mu=2\) and \(5\lambda+2\mu=1\).
Step 3: Solve for lambda and mu.
The two equations give \(\lambda=\frac{1}{19}\) and \(\mu=\frac{7}{19}\). Check: \(3\cdot\frac{1}{19}+5\cdot\frac{7}{19}=\frac{38}{19}=2\) and \(5\cdot\frac1{19}+2\cdot\frac{7}{19}=\frac{19}{19}=1\). Both hold, and both numbers are positive.
Step 4: Get the upper limit.
So \(2x+y\le \frac{1}{19}(15)+\frac{7}{19}(10)=\frac{85}{19}\). Then \(Z=19(2x+y)\le 85\).
Step 5: Show it is reached.
Equality needs both constraints to be tight, which happens at \(\left(\frac{20}{19},\frac{45}{19}\right)\). At that point \(Z=85\). So the optimum is 85.
Final Answer:
So the optimum value of Z is 85.
\[ \boxed{85} \]