Step 1 :Find the derivative of \(C(t) = 20t e^{-0.05t}\) using the product rule and the chain rule, which gives \(C'(t) = 20 e^{-0.05t} - t e^{-0.05t}\).
Step 2 :Set \(C'(t)\) equal to zero and solve for \(t\), which gives \(20 e^{-0.05t} = t e^{-0.05t}\) and \(t = 20\) minutes.
Step 3 :Substitute \(t = 20\) into \(C(t)\) to find the peak concentration, which gives \(C(20) = 20*20 e^{-0.05*20} = 326.9 \mathrm{ng/ml}\). So, \(\boxed{326.9 \mathrm{ng/ml}}\) is the peak concentration.
Step 4 :Substitute \(t = 15\) and \(t = 60\) into \(C(t)\) to find the concentration of the drug after 15 minutes and 60 minutes, which gives \(C(15) = 20*15 e^{-0.05*15} = 247.6 \mathrm{ng/ml}\) and \(C(60) = 20*60 e^{-0.05*60} = 43.3 \mathrm{ng/ml}\). So, the concentration after 15 minutes is \(\boxed{247.6 \mathrm{ng/ml}}\) and the concentration after 60 minutes is \(\boxed{43.3 \mathrm{ng/ml}}\).
Step 5 :Set \(C(t) = 10\) and solve for \(t\) to find when the next dose should be administered, which gives \(10 = 20t e^{-0.05t}\). Using a numerical method (like the Newton-Raphson method) gives \(t \approx 89\) minutes. So, the next dose should be administered in approximately \(\boxed{90}\) minutes (rounded to the nearest 10 minutes).