Download Midterm exam #1 - Han

MATH 1109 R12 MIDTERM EXAM 1 SOLUTION
SPRING 2014 - MOON
Write your answer neatly and show steps.
Except calculators, any electronic devices including cell phones are not allowed.
(1) Quick survey.
(a) (1pt) This class is:
Too easy
1
2
3
Moderate
4
5
6
Too difficult
7
(b) (1pt) Write any suggestion for improving this class. (For instance, give
more examples in class, explain proofs of formulas in detail, give more
homework, slow down the tempo, ...)
(2) (5 pts) The table gives the number of pairs of bald eagles in the continental
United States for various years, according to the US Fish and Wildlife Service.
Year
1988 1992 1996 2000
Number of pairs 2475 3749 5094 6471
Find the average rate of change of the number of pairs of bald eagles from 1992
to 2000.
average rate of change =
6471 − 3749
= 340.25
2000 − 1992
340.25 pairs per year
6471 − 3749
for the average rate of change: 3 pts.
2000 − 1992
• Getting the answer 340.25: 4 pts.
• Writing the answer with an appropriate unit 340.25 pairs per year: 5 pts.
• Stating the formula
Date: February 21, 2014.
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MATH 1109 Midterm Exam 1
Spring 2014 - Moon
(3) Let f (x) = x3 − 6x2 + 16.
(a) (5 pts) Find f 0 (x).
d 3
d
d
f 0 (x) =
x − 6 x2 + 16 = 3x2 − 6 · 2x + 0 = 3x2 − 12x
dx
dx
dx
• Evaluating the derivative f 0 (x) = 3x2 − 12x: 5 pts.
(b) (5 pts) Find the interval that f (x) is decreasing.
f (x) is decreasing when f 0 (x) < 0.
f 0 (x) = 3x2 − 12x = 3x(x − 4)
If 0 < x < 4, then x > 0 and x − 4 < 0 so f 0 (x) = 3x(x − 4) < 0. Therefore
on (0, 4), f (x) is decreasing.
• Factoring f 0 (x) and getting 3x(x − 4): 2 pts.
• Obtaining the interval (0, 4) correctly: 5 pts.
(c) (5 pts) Find the equation of the tangent line to y = f (x) at the point (2, 0).
f 0 (x) = 3x2 − 12x ⇒ f 0 (2) = 3 · 22 − 12 · 2 = −12
Equation of the tangent line:
y − y1 = m(x − x1 )
y − 0 = −12(x − 2)
y = −12x + 24
• Finding the slope −12 of the tangent line: 3 pts.
• Stating the equation of the tangent line y − y1 = m(x − x1 ): + 1 pt.
• Getting the equation y = −12x + 24: 5 pts.
(d) (5 pts) Find the interval that f (x) is concave upward. f (x) is concave upward if f 00 (x) > 0.
f 00 (x) = 6x − 12
f 00 (x) > 0 ⇔ 6x − 12 > 0 ⇔ x > 2
Therefore f (x) is concave upward on (2, ∞) (or x > 2).
• Getting the second derivative f 00 (x) = 6x − 12: 2 pts.
• Finding the interval x > 2: 5 pts.
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MATH 1109 Midterm Exam 1
Spring 2014 - Moon
√
(4) The equation of motion of a bird is g(t) = 5et + 12 t where g(t) is measured in
yards and t is the time in seconds.
dg
(a) (6 pts) Find .
dt
dg
d
1d 1
1 1 1
1 1
= 5 et +
t 2 = 5et + · t− 2 = 5et + t− 2
dt
dt
2 dt
2 2
4
dg
1 1
• Evaluating the derivative
= 5et + t− 2 : 6 pts.
dt
4
• For each computational mistake: -3 pts.
(b) (3 pts) Compute the velocity of the bird after 2 seconds.
velocity =
dg
1 1
|t=2 = 5e2 + 2− 2 ≈ 37.122
dt
4
37.122 yards per second
1 1
• Finding the velocity 5e2 + 2− 2 : 2pts.
4
• Writing the answer 37.122 yards per second with appropriate unit: 3
pts.
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MATH 1109 Midterm Exam 1
Spring 2014 - Moon
(5) (8 pts) For f (x) = x2 + 3x, evaluate f 0 (x) by using the definition of the derivative.
(In this problem, if you use differentiation rules, you cannot get any credit.)
f (x + h) − f (x)
h→0
h
((x + h)2 + 3(x + h)) − (x2 + 3x)
lim
h→0
h
2
2
x + 2xh + h + 3x + 3h − x2 − 3x
lim
h→0
h
2
2xh + h + 3h
lim
h→0
h
lim 2x + h + 3 = 2x + 0 + 3 = 2x + 3
f 0 (x) = lim
=
=
=
=
h→0
f (x + h) − f (x)
of f 0 (x): 2 pts.
h→0
h
((x + h)2 + 3(x + h)) − (x2 + 3x)
• Substituting f (x) = x2 +3x correctly and getting lim
:
h→0
h
4 pts.
• Expanding given formula, removing the common factor and obtaining lim 2x+
• Writing the definition lim
h→0
h + 3: 7 pts.
• Getting the correct answer 2x + 3: 8 pts.
• For each inappropriate use of mathematical notation: -2 pts.
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MATH 1109 Midterm Exam 1
Spring 2014 - Moon
(6) A food maker estimates that it costs
C(q) = 375 + 8q + 0.02q 2
dollars each day to produce q lunch boxes.
(a) (8 pts) How many boxes should be made daily in order to minimize the
average cost?
The minimum average cost occurs when the marginal cost is equal to the
C(q)
average cost. In other words, C 0 (q) =
.
q
C(q)
375 + 8q + 0.02q 2
0
C (q) = 8 + 0.04q,
=
q
q
C 0 (q) =
C(q)
375 + 8q + 0.02q 2
⇒ 8 + 0.04q =
q
q
375 + 8q + 0.02q 2 = q(8 + 0.04q) = 8q + 0.04q 2
375 + 0.02q 2 = 0.04q 2
375
0.02q 2 = 375 ⇒ q 2 =
0.02
r
375
q=
≈ 136.93
0.02
137 boxes
C(q)
for minimum average cost: 2 pts.
q
Finding C 0 (q) = 8 + 0.04q: 3 pts.
375 + 8q + 0.02q 2
Making the equation 8 + 0.04q =
for minimum avq
erage cost: 4 pts.
r
375
Solving the equation and getting q =
: 7 pts.
0.02
Rounding the answer and getting the answer 137 boxes: 8 pts.
• Stating the condition C 0 (q) =
•
•
•
•
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MATH 1109 Midterm Exam 1
Spring 2014 - Moon
(b) (8 pts) The food maker estimates that the expected revenue is given by
R(q) = 48q − 0.012q 2
dollars. Determine the number of boxes that the food maker should prepare in order to maximize profit.
R0 (q) = 48 − 0.024q
R0 (q) = C 0 (q) ⇒ 48 − 0.024q = 8 + 0.04q
40
= 625
0.064
625 boxes
0.064q = 40 ⇒ q =
•
•
•
•
Stating the condition R0 (q) = C 0 (q): 2 pts.
Finding R0 (q) = 46 − 0.028q: 3 pts.
Setting the equation 46 − 0.028q = 6 + 0.04q: 4 pts.
Solving the equation and getting the answer 625 boxes: 8 pts.
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