Download Mid-Term Exam 3 Solutions 1 Exam Solutions

MATH 151.01—Calculus and Analytic Geometry I
Mid-Term Exam 3 Solutions
March 9, 2012
Sam Henke
[email protected]
1
Exam Solutions
Problem 1
A paper cup has the shape of a cone with height 5 in. and a radius of 2 in. at the top. If water is poured into
3
the cup at a rate of 0.5 ins , how fast is the water rising when the water is 3 in. deep?
The water forms a smaller cone which is similar (in the geometric sense) to the paper cup. Let r be
the radius of the water in inches, let h be the height of the water in inches, let V be the volume of
the water in cubic inches, and let t be time in seconds. We’re given
.
know dh dV
dt
3
= 0.5 ins , and we want to
dt h=3
1 2
πr h
3
2
r= h
5
2
2
4
1
h h=
πh3
V= π
3
5
75
dV
4
dh
=
πh2
dt
25
dt
4
dh
36π
dh
0.5 =
π(3)2
=
25
dt h=3
25 dt h=3
dh 25 in
in
=
≈ 0.1105
dt h=3 72π s
s
V=
1
volume of a cone
similar triangles
substitute
differentiate both sides
evaluate at h = 3
solve
Problem 2
Let f(x) = 8x3 + 12x2 − 144x + 55 on the domain [−6, 4].
(a) Determine all critical numbers of f on the open interval (−6, 4).
(b) Determine all intervals where f is increasing, and where it is decreasing.
(c) Determine all intervals where f is concave up, and where is it concave down.
(d) Determine all inflection points of f.
(e) Determine the absolute maximum and minimum of f on its domain [−6, 4].
(a) Note that f is a polynomial, so it’s differentiable everywhere. Thus, the only critical points are
at points where the derivative is zero. We also only want the critical points on the interval
given, although that makes no difference in this case.
f 0 (x) = 24x2 + 24x − 144 = 0
differentiate and set equal to zero
24(x2 + 1 − 6) = 0
factor out 24
24(x + 3)(x − 2) = 0
factor
x = 2, −3
solve
(b) We must determine the signs of f 0 in different intervals.
f 0 (x) = 24(x + 3)(x − 2)
+
−∞
0
0
−
−3
+
from (a)
∞
consider signs in each interval
2
So f is increasing on [−6, −3) and (2, 4], and f is decreasing on (−3, 2).
(c) We must determine the signs of f 00 in different intervals.
f 0 (x) = 24x2 + 24x − 144
from (a)
f (x) = 48x + 24 = 24(2x + 1)
differentiate
00
−∞
−
0
+
∞
consider signs in each interval
1
−2
So f is concave up on (− 12 , 4], and f is concave down on [−6, − 21 ).
(d) There is an inflection point at − 12 by part (c).
2
(e) We check the value of f at critical points and at the end points.
f(−6) = 8(−6)3 + 12(−6)2 − 144(−6) + 55 = −377
f(−3) = 8(−3)3 + 12(−3)2 − 144(−3) + 55 = 379
f(2) = 8(2)3 + 12(2)2 − 144(2) + 55 = −121
f(4) = 8(2)3 + 12(2)2 − 144(2) + 55 = 183
The absolute maximum of f is on [−6, 4] is at (−3, 379), and the absolute minimum is at
(−6, −377).
Problem 3
Come ask me if you want a sketch of this graph.
Problem 4
Determine whether the statement is true or false. Justify your answer.
(a) If f has an absolute minimum at c, then f 0 (c) = 0.
(b) If f is a function for which both f 0 and f 00 exist for all real numbers, and f 0 (x) is always nonzero, then
f(2) 6= f(5).
(a) This is false. One reason is that an absolute minimum can occur at an end point—not just
a local minimum—so there’s no reason the derivative must be 0. Another reason is that an
absolute minimum does not have to have derivative 0; if the derivative exists, then it must be
0, but the derivative does not have to exist. For example, the absolute value function has an
absolute minimum at (0, 0), but the derivative there does not exist.
(b) This is true. Because f 00 exists for all real numbers, f 0 must be continuous for all real numbers. If f 0 were sometimes negative and sometimes positive, then by the Intermediate Value
Theorem there must be a point c such that f 0 (c) = 0. But f 0 is nonzero everywhere, so this is
a contradiction. Hence f 0 is either always positive or always negative; i.e., f is either always
strictly increasing or always strictly decreasing. In either case f(2) 6= f(5).
Another proof involves Rolle’s Theorem, which is a special case of the Mean Value Theorem.
By assumption, f 0 is defined for all real numbers, so f is continuous and differentiable everywhere. Thus, the hypotheses of Rolle’s Theorem applies. If f(2) = f(5), then there would
be some c ∈ (2, 5) such that f 0 (c) = 0. But again, f 0 is nonzero for everywhere, so this is a
contradiction. Therefore f(2) 6= f(5).
3
Problem 5
Find two positive integers such that the sum of the first number and four times the second number is 2000,
and the product of the numbers is as large as possible.
Let x be the first number, let y be the second number, and let P = xy be their product.
x + 4y = 2000
x
y = 500 −
4
x2
x
= 500x −
P = x 500 −
4
4
dP
x
= 500 − = 0
dx
2
x = 1000
y=
from problem statement
solve for y
substitute
differentiate to find critical points
solve for x
1000
= 250
4
substitute and solve for y
At this point, we know only that x = 1000 is a critical point—not necessarily an absolute maximum
or even a local maximum. We must use the First or Second Derivative Test for Absolute Extreme
Values. Note that P 0 (x) > 0 for x < 1000 and P 0 (x) < 0 for x > 1000, so that x is an absolute
maximum by the First Derivative Test for Absolute Extreme Values. Alternatively, P 00 (x) = − 41 is
negative for all values of x, so that P is concave down everywhere, so x = 1000 gives an absolute
maximum by the Second Derivative Test for Absolute Extreme Values. If you didn’t do one of these
steps, then you should have lost two points.
4