C(t)=0.32t/(t+2)².
C’(t)=0=(0.32(t+2)²-0.32t×2(t+2))/(t+2)⁴ at a maximum or minimum.
So 0.32(t+2)²=0.64(t²+2t), t²+4t+4=2t²+4t, 4=t² so t=2hr.
When t=2, C=0.64/16=0.04mg if C is the dose in mg. t=2 is a maximum because when t=0, C=0 and when t=3, C=0.96/25=0.0384 which is less than 0.04.
Solution: Drug is at its maximum at 2 hours and the dose is 0.04mg.