MATH 1109 R13 MIDTERM EXAM 2 SOLUTION
SPRING 2014 - MOON
Write your answer neatly and show steps.
Except calculators, any electronic devices including cell phones are not allowed.
(1) Differentiate the functions.
(a) (5 pts) f (x) = x2 ln x
d 2
d
1
f 0 (x) =
x · ln x + x2 ·
ln x = 2x ln x + x2 · = 2x ln +x = x(2 ln x + 1)
dx
dx
x
d 2
d
x · ln x + x2 ·
ln x: 2 pts.
dx
dx
• Obtaining the answer x(2 ln x + 1): 5 pts.
• Applying the product rule and getting
(b) (5 pts) y = (x2 + 5)100
u = x2 + 5 ⇒ y = u100
dy du
dy
=
·
= 100u99 · 2x = 100(x2 + 5)99 2x = 200x(x2 + 5)99
dx
du dx
• Finding an appropriate substitution u = x2 + 5 and y = u100 : 2 pts.
• Getting the answer 200x(x2 + 5)99 : 5 pts.
Date: April 4, 2014.
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MATH 1109 Midterm Exam 2
Spring 2014 - Moon
(2) A zombie population is increasing at a constant relative growth rate in a community. It is estimated that after three days the population increased from 4,000
to 8,200.
(a) (7 pts) Find a formula for the zombie population after t days.
The general population formula is P (t) = Cekr where C is the initial population. So we have
P (t) = 4000ekt .
Because P (3) = 8200,
8200
82
8200 = 4000e3k ⇒ e3k =
=
4000
40
1 82
82
⇒ k = ln
≈ 0.239
⇒ 3k = ln
40
3 40
P (t) = 4000e0.239t
• Stating the general population formula P (t) = Cekt : 2 pts.
• By using the initial population, getting P (t) = 4000ekt : 3 pts.
• Finding k ≈ 0.201 from P (3) = 8200 and obtaining P (t) = 4000e0.239t :
7 pts.
(b) (3 pts) What is the estimated zombie population after ten days? (Round the
answer to the nearest whole number.)
P (10) = 4000e0.239·10 ≈ 43654.
43654 zombies.
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MATH 1109 Midterm Exam 2
Spring 2014 - Moon
x4
− x3 + 2.
4
(a) (3 pts) Find all critical numbers.
(3) Let f (x) =
f 0 (x) = x3 − 3x2 = x2 (x − 3)
x2 (x − 3) = 0 ⇔ x = 0, 3
• Evaluating the derivative f 0 (x) = x2 (x − 3): 1 pt.
• Finding all critical numbers x = 0, 3: 3 pts.
(b) (4 pts) Find all local maxima and local minima.
x<0 0 0<x<3 3 x>3
x
f (x)
−
0
−
0
+
So f (3) = −4.75 is a local minimum and f (0) is not a local maximum nor
minimum.
• Drawing the table of the derivative: 2 pts.
• Making the conclusion that f (3) = −4.75 is a local minimum: 4 pts.
0
(c) (4 pts) Find the interval that y = f (x) is concave downward. y = f (x) is
concave downward ⇔ f 00 (x) < 0.
f 00 (x) = 3x2 − 6x = 3x(x − 2)
3x(x − 2) = 0 ⇔ x = 0, x = 2
3x(x − 2) < 0 ⇔ 0 < x < 2
Answer: (0, 2).
• Evaluating the second derivative f 00 (x) = 3x2 − 6x: 1 pt.
• Finding two zeros 0, 2 of f 00 (x): 2 pts.
• Obtaining the interval (0, 2): 4 pts.
(d) (4 pts) Find the absolute maximum and minimum on the interval [−1, 4].
f (−1) = 3.25, f (0) = 2, f (3) = −4.75, f (4) = 2
The absolute maximum is f (−1) = 3.25, and the absolute minimum is
f (3) = −4.75.
• Evaluating the values at critical numbers and two endpoints: 3 pts.
• Getting the correct answer (the absolute maximum is 3.25, the absolute minimum is -4.75): 4 pts.
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MATH 1109 Midterm Exam 2
Spring 2014 - Moon
(4) (6 pts) Find the derivative
dy
if x2 + y 2 = xy.
dx
x2 + y 2 = xy
dy
dy
=1·y+x·
2x + 2y
dx
dx
dy
dy
2x + 2y
=y+x
dx
dx
dy
dy
−x
= y − 2x
2y
dx
dx
dy
(2y − x)
= y − 2x
dx
dy
y − 2x
=
dx
2y − x
d
dy
dy
• By taking the derivative
of both side, obtaining 2x + 2y
= y+x : 4
dx
dx
dx
pts.
dy
y − 2x
dy
and getting
=
: 6 pts.
• Solving it for
dx
dx
2y − x
(5) (8 pts) By using calculus, find the dimensions of a rectangle with perimeter 80 m
whose area is as large as possible. Let x and y be two sides of the rectangle.
Because its perimeter is 80 m, 2x + 2y = 80.
2x + 2y = 80 ⇒ 2y = 80 − 2x ⇒ y = 40 − x
We want to maximize the area, which is A = xy = x(40 − x) = 40x − x2 .
Because it is a length of a line segment, x ≥ 0. Also from 0 ≤ y = 40 − x,
x ≤ 40. So we have 0 ≤ x ≤ 40.
dA
= 40 − 2x
dx
40 − 2x = 0 ⇔ x = 20
A(0) = 0, A(10) = 20 · 10 − 102 = 100, A(20) = 0
When x = 20, y = 40 − 20 = 20. Therefore the maximum area is 400 m2 and it
occurs when both sides are 20 m.
• Finding a relation between two sides 2x + 2y = 80: 2 pts.
• Expressing the area function A = 40x − x2 with respect to x: 4 pts.
• Taking the derivative of A and finding the critical number 20: 6 pts.
• Comparing the values at the critical numbers and two endpoints: 7 pts.
• Getting the dimensions (both sides are 20 m): 8 pts.
4
MATH 1109 Midterm Exam 2
Spring 2014 - Moon
(6) A manufacturer of utility trucks estimates that the demand function for their
smallest model truck is p = 58000 − 32q. They currently charge $28, 000 for each
truck.
(a) (4 pts) What is the elasticity of demand at this price point? (Round to two
decimal places.)
E(q) = −
D(q)
qD0 (q)
D(q) = 58000 − 32q ⇒ D0 (q) = −32
58000 − 32q
58000 − 32q
E(q) = −
=
q(−32)
32q
58000 − 28000
p = 28000 ⇒ 28000 = 58000 − 32q ⇒ q =
≈ 937.5
32
58000 − 32 · 937.5
≈ 0.93
E(937.5) =
32 · 937.5
• Writing the elasticity formula −
D(q)
: 1 pt.
qD0 (q)
58000 − 32q
: 2 pts.
32q
• Obtaining the quantity q = 937.5 when p = 22000: 3 pts.
• Finding elasticity 0.93 at the point: 4 pts.
• By using D(q) = 58000 − 32q, finding E(q) =
(b) (3 pts) To collect more revenue, should the manufacturer raise prices or
lower prices for this truck model?
Because E(q) < 1, the manufacturer should raise the price.
• If the elasticity in (a) is not correct, one cannot get the full credit.
(c) (4 pts) What price gives the maximum revenue? (Round to two the nearest
cents.) The maximum revenue occurs when E(q) = 1.
E(q) = 1 ⇔
58000 − 32q
58000
= 1 ⇒ 58000 − 32q = 32q ⇒ q =
≈ 906
32q
64
p = 58000 − 32 · 906 = 29008
Answer: $29008
• Setting up an equation
58000 − 32q
= 1 for the maximum revenue: 1
32q
pt.
• Solving the equation and getting q ≈ 906: 3 pts.
• Getting the price $29008 maximizes the revenue: 4 pts.
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