To determine
To find: the maximum beverage BAC during the first 3 hours
Answer
C(0.36)=0.177 and C(3)=9.05×10−4
Explanation
Given C(t)=1.35t e−2.802t
C(t)=1.35t e−2.802tDifferenetiating with respect to tC′(t)=1.35[t.e−2.802t.(−2.802)+e−2.802t]
For maxima
C′(t)=0⇒1−(2.802)t=0⇒t=12.802=0.3569≈0.36⇒t=21.4min
Therefore C(0.36)=1.35(0.36) e−2.802(0.36)=0.177
C(3)=1.35(3) e−2.802(3)=9.05×10−4
Conclusion: C(0.36)=0.177 and C(3)=9.05×10−4