It gradually decreases as alcohol is metabolized. The function C(t)=0.135 t e^{-2.802 t}models the mean BAC measured in g/mL.
The maximum average BAC during 3 hours is 0.0001358 g/mL.
f(t) = α t e−βt --(1)
Let's rewrite this in a simple form:
f(t)= α eˡⁿ ᵗ e⁻βt = αe^(ln t −βt)
Since e^x is strictly increasing and it will be maximized exactly when its argument is maximized, so we can maximize instead:
g(t)=ln t −βt
differentiating with respect to t , and g'(t) = 0
g′(t)=1/t − β = 0
=> t =1/β
we have given a function
C(t)=0.135 t e⁻²·⁸⁰²ᵗ
if we compare it with (1) we get
β = 2.802, 0.135 = α
For it's maximized we need to check the second order condition, and that of g will differentiate again , g′′(t)= −1/t² < 0
We have to compute the derivative of C(t):
C′(t) = 0.135 t⋅(−2.802)e⁻²·⁸⁰²ᵗ + 1.35e⁻²·⁸⁰²ᵗ
For optimum at t₀ if C′(t₀)=0 and C′′(t₀)≠0. Here, we have
C′(t₀) = 0.135t₀⋅(−2.802)e⁻²·⁸⁰²ᵗ₀+ 0.135e⁻²·⁸⁰²ᵗ₀ =e⁻²·⁸⁰²ᵗ₀(−0.135* 2.802t₀+ 0.135)=0
It is clear that e⁻²·⁸⁰²ᵗ₀ not equal to zero for all t₀∈R, so that
=> −0.135* 2.802t₀+0.135=0
=> t₀ = 1/2.802 ≈0.36
let us consider t is in hours, so that it makes t₀ =0.36h≈21.41min. This is the only optimum and one should verify it is indeed a maximum, i.e. C′′(t₀)<0.
Now, easily compute the maximum average BAC, which is C(t₀)=C(0.36) = 0.135 (0.36)e⁻²·⁸⁰²⁽⁰·³⁶⁾
= 0.0486(0.3678) = 0.01787508
Hence, the maximum average BAC, is 0.017 g/dL.
Maximum average BAC during the first 3 hours,
t = 3 , C(t)=C(3) = 0.135 (3)e⁻²·⁸⁰²⁽³⁾ = 0.0001358 g/mL
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