Exam A
Exam 3
1.(6 pts) The function f (x) = x3 − 2x2 has:
(a) Two critical points; one is a local minimum, the other is
neither a local maximum nor a local minimum.
(b) Two critical points; one is a local maximum, the other is
neither a local maximum nor a local minimum.
(c) Two critical points, both local maxima.
(d) Two critical points, both local minima.
(e) Two critical points; one is a local maximum, the other a local minimum.
2.(6 pts) For a continuous function f (x), consider the following conditions:
• f 0 (x) > 0 on (−2, −1) ∪ (0, 1).
• f 0 (x) < 0 on (−∞, −2) ∪ (−1, 0) ∪ (1, 2) ∪ (2, +∞).
Which of the following statements must be true, for any function satisfying the above
conditions?
(a) The function has at least 4 critical points, two of which must be local maxima,
and two must be local minima.
(b) The function has at most 4 critical points, which could be local maxima or minima.
(c) The function has at most 3 critical points, one of which is a local minimum,
the others local maxima.
(d) The function has at most 3 critical points, and no local maximum.
(e) The function has no critical points, since it is continuous everywhere.
Exam A
x
, restricted to the
+1
interval [0, 2], attains a global maximum. In other words, f (m) > f (x) for all x ∈ [0, 2].
Which of the following statements about m is correct?
3.(6 pts) Let m ∈ [0, 2] be the point where the function f (x) =
(a) m = 2.
(b) 1/2 < m < 3/2.
(d) 0 < m < 1/4.
(e) 1/4 < m < 1/2.
x2
(c) 3/2 < m < 2.
4.(6 pts) Suppose that a company determines that given a price p per unit of a commodity,
the number of units sold q is q = 20e−0.5p . Determine the maximal revenue R = pq.
(a) 20e−0.75
(b) 40e−1.0
(c) 30e−1.0
(d) 20e−0.25
(e) 10e−0.5
Exam A
5.(6 pts) Consider the curve given by
x2 + y 3 + 2xy + xy 2 = 17.
Determine
(a)
dy
at the point (1, 2).
dx
1
2
(b) − 12
(c) − 95
(d)
5
9
(e) 0
6.(6 pts) Determine all of the inflection points for the function f (x) = x4 − 2x3 − 12x2 − 5.
(a) x = −2
(b) x = 2, x = −1
(d) x = −2, x = 1
(e) f does not have any inflection points.
(c) x = 1
Exam A
7.(6 pts) The maximum value of the function y = x3 − 3x − 1 on the interval [0, 3] is
(a) −1
(b) −3
(c) 1
(d) 17
(e) 15
8.(6 pts) Find the x-coordinate of all points where the graph of y = 3x2 + 6x + 3
has slope 2.
3
and x = −1
5
√
√
(c) x = 1 + 2 and x = 1 − 2
(a) x =
(e) x = 0 and x = −1
(b) There are no points where the slope is 2.
(d) x = −
2
3
Exam A
9.(6 pts) A restaurant owner studied the sales of an octopus dish and determined that its
72
, where p is the price in dollar of
average number of orders q each night is given by p =
q+2
an order of the dish. Supposing each appetizer costs the restaurant $4 to make, what price
should the owner charge to maximize profit from the appetizer?
(a) 12
(b) 8
(c) 3
(d) 14
(e) The more he charges the greater the profit.
10.(6 pts) Consider the function f (x) =
(a) (0.5, 3)
(b) (−1, 3)
(e) f is increasing on its domain.
1 + ln x
. On which interval below is f decreasing?
x
(c) (2, 5)
(d) (0.5, 5)
Exam A
11.(12 pts) A cylindrical tank is filling with water at a rate of 4t cubic meters per minute at
any given time t. Assume that the radius of the tank is 2 meters.
(1) Determine the rate at which the height of the water is changing at any given time t.
(2) Determine the rate at which the height of the water is changing at time t = 3.
12.(14 pts) A certain company can produce up to 500 units of a product it wants to sell. The
demand function for this product is p = 1000e−q/100 , where p is the price in dollars and
q is the quantity being produced. Find the production level (i.e., how many units) that
maximizes the profit of this company, knowing that it has a fixed cost of $700.
This problem is badly phrased. From the solution we see that it was assumed that $700 is
the cost no matter what the level of production. This is not usually what fixed cost means.
Exam A
13.(14 pts) NASA is tracking an asteroid on a parabolic orbit passing close to earth. In their
coordinate system the earth is at the point (0, 10) and the orbit is the graph of y = x2 . How
close does the asteroid come to the earth?
Be sure to explain why your distance is a minimum and not just a local minimum (or even
worse, a local maximum).
Hint: Minimizing the distance squared makes your calculations easier.
Exam A
1. Solution. The critical points are solutions to f 0 (x) = 3x2 − 4x = 0, which are x = 0 and
x = 4/3. Since f 00 (x) = 6x − 4, we have f 00 (0) = −4 < 0 and f 00 (4/3) = 4 > 0, so by the
second derivative test, one is a local maximum and the other a local minimum.
2. Solution. The conditions given imply that f 0 (−2) = f 0 (−1) = f 0 (0) = f 0 (1) = 0, but
we cannot say what happens to f 0 (2); it could be negative or zero, so we could have up to
5 critical points. This information alone determines the right answer. By looking at the
signs of f 0 (x) on the given intervals, we can easily see that twice it changes from positive to
negative, and twice it changes from negative to positive, so we have two local maxima, two
local minima.
3. Solution. First, find the critical points on the interior of [0, 2] by solving the equation
(1)
f 0 (x) =
1(x2 + 1) − x(2x)
1 − x2
=
(x2 + 1)2
(x2 + 1)2
1 − x2
= 0 ⇒ x = ±1.
(x2 + 1)2
In [0, 2], x = 1 is the only critical. Now, comparing the values of f (x) at the critical point
x = 1 with the values at the endpoints, we see that f (1) = 1/2 is a global maximum. Thus,
m = 1 and 1/2 < m < 3/2.
(2)
f 0 (x) = 0 ⇒
Exam A
4. Solution. Taking the first derivative of R(p) = 20pe−0.5p , we have
R0 (p) = 20e−0.5p − 10pe−0.5p .
Setting this equal quantity equal to zero,
20e−0.5p − 10pe−0.5p = 0
=⇒
p = 2.
Choosing values to the left and right of p = 2, in this case p = 0 and p = 100, we have R0 (0)
positive and R0 (100) negative. Hence we see that p = 2 is a maximum. Plugging this value
back into our function gives our maximal revenue as
R(2) = 40e−1
5. Solution. Using implicit differentiation, we see that
dh i
dh 2
x + y 3 + 2xy + xy 2 ] =
6
dx
dx
dy
dy
dy
+ 2y + 2x + y 2 + 2xy
=0
=⇒ 2x + 3y 2
dx
dx
dx
dy
dy
dy
=⇒ 2(1) + 3(2)2
+ 2(2) + 2(1) + (2)2 + 2(1)(2)
=0
dx
dx
dx
dy
=⇒ 18
= −10
dx
dy
5
=⇒
=−
dx
9
Exam A
6. Solution.
We see that the first derivative for f is
f 0 (x) = 4x3 − 6x2 − 24x
and the second derivative is
f 00 (x) = 12x2 − 12x − 24
Setting the second derivative equal to zero we find
12x2 − 12x − 24 = 0
=⇒
x2 − x − 2 = 0
=⇒
=⇒
(x − 2)(x + 1) = 0
x = 2, x = −1
Checking the three values
• x = −10, f 00 (−10) = (12)(−10 − 2)(−10 + 1) = (12)(−12)(−9) > 0.
• x = 0, f 00 (0) = (12)(0 − 2)(0 + 1) = (12)(−2)(1) < 0.
• x = 10, f 00 (10) = (12)(10 − 2)(10 + 1) = (12)(8)(11) > 0.
+
-1
+
2
We see that the sign switches at each zero so that x = −2 and x = 1 are both inflection
points.
7. Solution. Find critical points: y 0 = 3x2 − 3; y 0 = 0 has solutions ±1. y(0) = −1; y(1) = −3;
y(3) = 17.
8. Solution. y 0 = 6x + 6. Solve 6x + 6 = 2, 6x = −4, x = −
2
3
Exam A
72q
72q
. Cost C = 4q so profit P =
− 4q =
q+2
q+2
2
(64
−
8q)(q
+
2)
−
(64q
−
4q
)(1)
2
2
72q − 4q(q + 2)
72q − 4q − 8q
64q − 4q
=
=
. P0 =
=
q+2
q+2
q+2
(q + 2)2
128 − 16q − 4q 2
32 − 4q − q 2
(8 + q)(4 − q)
64q + 128 − 8q 2 − 16q − 64q + 4q 2
=
=
4
=4
.
2
2
2
(q + 2)
(q + 2)
(q + 2)
(q + 2)2
Solutions are q = −8 and q = 4 and we can’t charge a negative price so q = 4 and hence
72
p=
= 12.
4+2
9. Solution. Revenue R = q · p =
( x1 ) · x − (1 + ln x) · (1)
1 − 1 − ln x
− ln x
10. Solution. f (x) =
=
=
. Solve f 0 (x) = 0 so
2
2
x
x
x02
ln x = 0 so x = 1 is the only critical point. If x > 1, ln x > 0 so f < 0; x < 1, ln x < 0 so
f 0 > 0.
0
−
+
0
1
so decreasing on any subinterval of (1, ∞)
Exam A
11. Solution.
(1) We see that
V = πr2 h
=⇒
=⇒
=⇒
so
V = 4πh since r = 2
dV
dh
= 4π
dt
dt
dh
4t = 4π
dt
dh
t
=
dt
π
(2)
dh
3
=
dt
π
12. Solution. The profit function, in terms of the quantity q being produced, will be
P (q) = R(q) − C(q) = pq − 700 = 1000qe−q/100 − 700.
To maximize it, first find its critical points:
−q/100
e
q
0
−q/100
P (q) = 1000 −
+e
= 0 ⇒ q = 100.
100
Comparing the values at the endpoints with the value at the critical point, we have
P (0) = −700
100, 000
(4)
P (100) =
− 700
e
500, 000
(5)
P (250) =
− 700 < 100, 000 − 700.
e5
Thus, the global maximum is achieved when q = 100.
(3)
13. Solution. Distance squared is D = r
x2 + (x2 − 10)2 = x2 + x4 − 20x2 + 100 = x4 − 19x2 + 100.
19
D0 = 4x3 − 38x so x = 0 and x = ±
≈ ±3.08 are the critical points.
2
r !
19
19
Since D00 = 12x2 −38, D00 (0) = −38, local max.; D00 ±
= 12 −38 = 114−38 > 0,
2
2
local minimum’s. You can also use the first derivative test.
Since lim = ∞ the asteroid orbit is eventually arbitrarily far away from the earth.
x→±∞
0
-
+
-
+
D
-3.08
0
3.08
Hence as the asteroid its distance to the earth is steadily decreasing until it gets to x =
−3.08. Then its distance starts to increase until x = 0 where it is as far away from the
earth as it has been since x = −3.08. Then it starts to get closer to the earth again until in
gets to x = 3.08, after which the distance begins to increase again and gets steadily bigger
thereafter. Just
the full number line would get you full marks.
r ! 2
19
19
19 · 19 − 2 · 19 · 19 + 400
400 − 192
400 − 361
19
D ±
=
=
=
=
−19 +100 =
2
2
2
4
4
4
r
39
39
so the minimum distance is
≈ 3.122.
4
4
Another approach. D = x2 + (y − 10)2 = y + (y − 10)2 .
There is one critical point at y =
dD
= 1 + 2(y − 10) = 2y − 19.
dy
19 d2 D
.
= 1 + 2(y − 10) = 2 > 0 so local minimum. Since
2 dy 2
y = x2 , y takes values in [0, ∞).
−
+
0
D
0
19/2
Hence D decreases from y = 0 to y q
= 19/2 and increasesq
thereafter. Hence 19/2 is a global
q
q
√
2
19
19
1 2
38
1
39
minimum and the distance is D = 19
+
−
10
=
+
−
=
+
=
.
2
2
2
2
4
4
4
dD
dy
Yet another approach. D = x2 + (y − 10)2 .
= 2x + 2(y − 10) . Since y = x2 ,
dx
dx
dy
dD
2
2
2
= 2x so
= 2x + 2(x − 10)2x = 2x(1 + 2x − 20) = 2x(2x − 19). Proceed from here
dx
dx
as in the first solution.