MATH 205 Final Exam Solutions Name: 1. (12 points) The keratoid ...

MATH 205 Final Exam Solutions Name:
Since f ′ (x) = 4− 3600 , we see that f ′ (x) will have 3 critical points: one when x = 0, where f ′ (x)
x 2
is undefined, and two at the solutions to 4 − 3600 = 0, which occur at x = ± √ 900 = ±30. Our
x 2
options for maximizing choices of x are thus the 3 critical points 0, −30, and 30, together with
the interval endpoints 0 and the limiting behavior as x → ∞. −30 is outside the interval and
may be rejected out of hand. Evaluating at the other three points, we see that f(x) does not
exist at x = 0, but that lim x→0 + f(x) = +∞; lim x→+∞ f(x) = +∞, and f(30) = 4 · 30 + 3600.
30
This last value need not be completely calculated; it is finite, and thus a better minimum than
the other two prospects. Thus, our area-minimizing choice of x is 30, which has an associated
value of y = 1200 = 40, so our optimal dimensions are 30 × 40.
30
3. (12 points) G-23 paxilon hydrochlorate is eliminated from the bloodstream at a rate of 12%
per hour. Miranda has just taken a 40mg intravenous dose.
(a) (4 points) Construct a function modeling the quantity of the drug in her body after t
hours.
Since the elimination rate is 12%, the quantity of the drug is f(t) = 40e −0.12t .
(b) (4 points) How quickly is the drug being eliminated after 2 hours?
The rate of eliminaiton is denoted by f ′ (t); here, f ′ (t) = 40·(−0.12)e −0.12t , and specifically
f ′ (2) = 40(−0.12)e −0.24 , which is approximately −3.77, i.e. an elimination rate of 3.77
mg/hour.
(c) (4 points) How many hours does it take for the quantity of the drug to drop below 10mg?
We solve for the value of t such that f(t) = 10:
10 = 40e −0.12t
1
4 = e−0.12t
ln 1 4 = −0.12t
ln 1 4
−0.12 = t
so there will only be 10mg remaining after ln 1 4
−0.12
4. (12 points) Let h(x) = (x 2 + 1)e x .
≈ 11.6 hours.
(a) (5 points) Where is h(x) increasing? Where is it decreasing?
We calculate h ′ (x) = (x 2 + 1)e x + 2xe x = (x 2 + 2x + 1)e x ; since h ′ (x) = (x + 1) 2 e x , it is
evident that h ′ (x) = 0 at x = −1, but is positive everywhere else, so h(x) is increasing
where h ′ (x) is positive, which is to say when x ≠ −1.
(b) (7 points) Determine h(x)’s concavity and identify points of inflection.
We calculate h ′′ (x) = (x 2 + 2x + 1)e x + (2x + 2)e x = (x 2 + 4x + 3)e x ; since h ′′ (x) =
(x + 1)(x + 3)e x , it is evident that h ′ (x) = 0 at x = −1 and x = −3. Now we can
probe in the three regions: when x < −3, when −3 < x < −1, and when x > −1, with
the three following respective calculations: h ′′ (−4) = ((−4) 2 + 4(−4) + 3)e −4 = 3e −4 ,
h ′′ (−2) = ((−2) 2 + 4(−2) + 3)e −2 = −e −2 , and h ′′ (0) = (0 2 + 4 · 0 + 3)e 0 = 3, so the
diagram of the sign of h ′′ (x) at various values is given by:
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