MATHEMATICS 184 (Section 921) - TERM EXAM #4
SOLUTIONS
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INSTRUCTIONS: Notes, books and calculators are not permitted. Answers may
receive no credit if accompanying work is not provided. When indicated, simplify
answers as much as possible, and be sure to clearly indicate the final answer for
each question. Please use the back of the pages if you need extra space.
9
1. A new company, ACME Re-fried Beans Inc. is trying to determine the most economical
cylindrical can in which to package their product. If the can must be cylindrical, and must
have a volume of 54π cm3 , what are the dimensions (height and radius) of the can that would
require the least material?
Hint: The cylinder is cut from sheet metal as 2 circles (top and bottom) and a rectangular
piece that is rolled up and glued end to end
This is asking us to minimize the surface area of the cylinder.
A = 2πrh + 2 · πr2
We need to eliminate one of the variables, we’ll use the given fact that
Volume = πr2 h = 54π cm3
Solve this for h, h = 54π
= 54r2
πr2
Substituting into the formula for surface area we have
54
) + 2πr2
2
r
108π
+ 2πr2
=
r
A = 2πr(
The domain is (0, ∞)
To minimize area, we take the derivative and find the critical points
A0 = − 108π
+ 4πr
r2
This exists everywhere on (0, ∞) (note r = 0 is not in the domain) Next, try setting A0 = 0.
0 = −
108π
+ 4πr
r2
108π
= 4πr
r2
108π
r3 =
4π
3
r = 27
r = 3
So r=3 is the only critical point. Now we compare the area at critical points and endpoints
as r → 0 , A → ∞
as r → ∞ , A → ∞
108π
when r = 3 , A =
+ 2π(9)
3
which is less than infinity
Therefore, the dimensions that require the least material are r = 3 cm and h =
Page 2
54
9
= 6 cm.
7
2. Find the global maximum and the global minimum (if they exist) of the function
y = 3x5 − 5x3 − 1
on the interval [− 12 , 2]. Provide both x and y coordinates.
First, find the critical points. y0 = 15x4 − 15x2 y’exists everywhere on [− 12 , 2] (so no CP’s from
this) Next, set y0 = 0
0 = 15x4 − 15x2
0 = 15x2 (x2 − 1)
0 = 15x2 (x + 1)(x − 1)
So x=0, x=1, or x=-1. Therefore the critical points are x=0 and x=1 (note that x=-1 is not in
the domain).
Compare the critical points and endpoints
if x = 0
if x = 1
1
if x = −
2
if x = 2
, y = −1
, y = 3 − 5 − 1 = −3
1
1
1
5 17
, y = 3(− )5 − 5(− )3 − 1 = 3(− ) + =
2
2
32
8 32
, y = 3(2)5 − 5(2)3 − 1 = 96 − 40 − 1 = 55
So the global maximum of 55 occurs at x = 2 and the global minimum of −3 occurs at x = 1.
Page 3
3. For the function
y=
2
1
x2 − 1
−1
(a) Determine the domain and the range
D = (−∞, −1), (−1, 1) and (1, ∞)
R = (−∞, −2) and (−1, ∞)
3
(b) Find the coordinates of all x and y-intercepts
x-intercepts: (y=0)
0 =
1
x2 − 1
−1
1 = x2 − 1
2 = x2
√
x = ± 2
√
√
So ( 2, 0) and (− 2, 0) are the x-intercepts.
y-intercepts: (x=0)
1
y = 0−1
− 1 = −2
So (0,-2) is the y-intercept
3
(c) Find the coordinates of all local extrema (indicate which are maxima and which are minima)
y0 = −1(x2 − 1)−2 (2x) = (x−2x
2 −1)2
y’ is undefined when x = ±1 but not in the domain
0 = (x−2x
2 −1)2 ⇒ x = 0, y = −2.
So (0, −2) is the only critical point. We’ll use the first derivative test to check if it is a
max or min.
if x < 0, y0 > 0, (the function is increasing)
if x > 0, y0 < 0, (the function is decreasing)
So by the first derivative test (0, −2) is a local maximum.
Page 4
4
(d) Determine the intervals of concavity (ie. On which intervals is the function concave up?
on which is it concave down?)
First, we need to know the locations of inflection points.
−2(x2 − 1)2 − (−2x)(2)(x2 − 1)(2x)
(x2 − 1)4
−2(x2 − 1) + 8x2
=
(x2 − 1)3
6x2 + 2
=
(x2 − 1)3
y00 =
Since y00 is never equal to zero (because 6x2 + 2 can’t equal zero) there are no inflection
points in this function.
So, we need only check the intervals created by the domain for concavity:
on
on
on
2
(−∞, −1), x2 − 1 > 0, so y00 > 0
(−1, 1), x2 − 1 < 0, so y00 < 0
(1, ∞), x2 − 1 > 0, so y00 > 0
(e) Identify any horizontal asymptotes
limx→∞ x21−1 − 1 = −1
limx→−∞ x21−1 − 1 = −1
So there is a horizontal asymptote at y = −1.
3
(f) Identify any vertical asymptotes
This function is undefined at x = −1 and x = 1, so
limx→−1− x21−1 − 1 = +∞, and limx→−1+ x21−1 − 1 = −∞
So there is a vertical asymptote at x = −1.
limx→1− x21−1 − 1 = −∞, and limx→1+ x21−1 − 1 = +∞
So there is a vertical asymptote at x = 1.
Page 5
concave up
concave down
concave up
2
(g) Using the information from parts (a)-(f), sketch the function y =
below.
1
x2 −1
− 1 on the grid
4
y
2
–4
–2
0
2
4
x
–2
–4
Page 6
5
4. (a) Use the local linearization of the function
f (x) = ln(2x + 1)
to estimate ln(1.01)
near x = 0
f (x) ≈ f (a) + f 0 (a)(x − a)
Here a = 0, so
f (0) = ln(1) = 0
1
f 0 (x) =
(2)
2x + 1
f 0 (0) = 2
Therefore,
ln(2x + 1) ≈ 0 + 2(x − 0)
ln(2x + 1) ≈ 2x
So, to estimate ln(1.01) we take x = 0.005 (then 2x + 1 = 1.01)
Thus, ln(1.01) ≈ 0.01.
(note, it turns out that ln(1.01) = 0.0099995033.., so this is a decent approximation)
2
(b) Compute the following limits:
i. limx→∞ e−x ln(x)
ln(x)
∞
→
x
x→∞ e
∞
= lim
1
x
x→∞ ex
= lim
1
x→∞ xex
= 0
= lim
Page 7
so use L’Hopital
3
ii. limx→0 sin(x)−x
x3
This goes to
0
0
so use L’Hopital
0
cos(x) − 1
→
2
x→0
3x
0
−sin(x)
0
= lim
→
x→0
6x
0
−cos(x)
= lim
x→0
6
−1
=
6
= lim
Page 8
so use L’Hopital again
so use it again