Question

C(t) = 1.35te−2.802t † models the average BAC, measured in mg/mL, of a group of eight male subjects t hours after rapid consumption of 15 mL of ethanol (corresponding to one alcoholic drink). What is the maximum average BAC during the first 5 hours? (Round your answer to three decimal places.)

Answer 1

Therefore The Maximum BAC is 0.1772 mg/mL, and it occurs

1/2.802 = 0.3569 hours after the consumption.

hours after the consumption 0.36 hours approximately 3 decimal places

Explanation:

Given Equation

C(t) = 1.35te−2.802t

C'(t) = d/dt(1.35te−2.802t)

C'(t) = 1.35e- 2.802t + 1.35te--2.802t * (- 2.802)

= 1.35e-2.802(1 - 2.802t)

Because e-2.802t > 0 ,

C'(t) = 0

1.35e-2.802t (1 - 2.802t) = 0

1 - 2.802t = 0

t = 1/2.802

C(1/2.802) = 1.35 (1/2.802)e-2.802

= 1.35 * 1/2.802 e-1

= 0.1772

C(0) = 1.35 * 0 * e0

= 0

C(3) = 1.35 * 3 * e-2.802*3

≈ 0.0009

Therefore The Maximum BAC is 0.1772 mg/mL, and it occurs

1/2.802 = 0.3569 hours after the consumption.

hours after the consumption 0.36 hours approximately 3 decimal places