MATH 1109 R13 MATH FOR BUSINESS: CALCULUS FINAL SOLUTION
SPRING 2014 - MOON
• Write your answer neatly and show steps.
• You may use your calculator, but you may not use the graphing function
nor the integration function. You have to provide all details of integral
computation. If not, you cannot earn any credit.
(1) (a) (5 pts) Find the derivative of f (x) = 3x5 − 9x2 + 5.
f 0 (x) = 3 · 5x4 − 9 · 2x + 0 = 15x4 − 18x
• Computing the derivative f 0 (x) = 15x4 − 18x: 5 pts.
(b) (5 pts) Find the derivative of
g(x) =
x3
.
ex
By quotient rule,
x
d x
d 3
e − x3 dx
x
e
3x2 ex − x3 ex
3x2 − x3
=
=
g 0 (x) = dx
(ex )2
(ex )2
ex
(i) By using quotient rule, getting the derivative g 0 (x) =
5 pts.
(ii) Making a mistake for the sign: -2 pts.
Date: May 7, 2014.
1
3x2 − x3
:
ex
MATH 1109 R13 Final
Spring 2014 - Moon
(c) (5 pts) Find the derivative of h(x) = e
√
x
.
√
y = h(x) = e x
√
1
u = x = x 2 ⇒ y = eu
h0 (x) =
dy
1 1 1
dy du
1 1
=
·
= e u · x− 2 = e x 2 x− 2
dx
du dx
2
2
1
• Defining an appropriate u = x 2 : 2 pts.
du
1 1
dy
= eu ,
= x− 2 : 4 pts.
• Computing two derivatives
du
dx
2
1 x 12 − 1
2
• Getting the answer e x : 5 pts.
2
(d) (5 pts) Find the equation of the tangent line to y = x ln x at (e, e).
1
dy dy
= 1 · ln x + x · = ln x + 1 ⇒
= ln e + 1 = 2
dx
x
dx x=e
By using the equation of line y −y1 = m(x−x1 ), we have the tangent
line
y − e = 2(x − e),
or
y = 2x − e.
dy
= ln x + 1: 2 pts.
• Calculating
dx
• Computing the slope 2 of the tangent line: 3 pts.
• Getting the equation of the tangent line y = 2x − e: 5 pts.
2
MATH 1109 R13 Final
Spring 2014 - Moon
(2) The following is the graph of y = f 0 (x).
(a) (2 pts) On what intervals is f increasing?
(−∞, 1), (5, 7)
(b) (2 pts) Find all critical numbers of f .
1, 3, 5, 7
(c) (2 pts) At what values of x does f have a local maximum?
A function f has a local maximum if the derivative changes from
positive to negative as x increases. Therefore f has a local maximum
at x = 1, 7.
(d) (2 pts) At what values of x does f have an inflection point?
A function f has an inflecton point if the sign of the second derivative (equivalently, the slope of f 0 ) does change. So there are four
values with inflection points, x = 2, 3, 4, and 6.
3
MATH 1109 R13 Final
Spring 2014 - Moon
(3) (8 pts) Car A is traveling west at 50 mi/h and car B is traveling north at
40 mi/h. Both are headed for the intersection of the two roads. At what
rate are the cars approaching each other when car A is 1.2 mi and car
B is 0.5 mi from the intersection? Find the precise answer or round the
answer to two decimal places.
Let x be the distance from the intersection to the car A, and let y be
the distance from the intersection to the car B. And let z be the distance
between two cars. By Pythagorean theorem, x2 + y 2 = z 2 . Also from the
assumption,
dx
dy
= −50,
= −40.
dt
dt
dz We want to find .
dt x=1.2,y=0.5
dx
dy
dy
x2 + y 2 = z 2 ⇒ 2x + 2y
= 2z
dt
p dt
pdt
When x = 1.2, y = 0.5, z = x2 + y 2 = (1.2)2 + (0.5)2 = 1.3. Therefore
2x
dy
dy
dz
dx
+ 2y
= 2z
⇒ 2 · 1.2 · (−50) + 2 · 0.5 · (−40) = 2 · 1.3
dt
dt
dt
dt
dz
dz
160
⇒ 2.6 = −160 ⇒
=−
≈ −61.54
dt
dt
2.6
160
answer: −
mi/h
2.6
• Finding a relation x2 + y 2 = z 2 between variables: 1 pt.
dy
dx
= −50,
= −40: 2 pts.
• Stating the conditions
dt dt
dz • Indicating the quantity we want to find: 4 pts.
dt x=1.2,y=0.5
dx
dy
dz
• By using implicit differentiation, finding 2x +2y
= 2z : 6 pts.
dt
dt
dt
160
• Getting the answer −
mi/h and writing it with an appropriate
2.6
unit: 8 pts.
4
MATH 1109 R13 Final
Spring 2014 - Moon
(4) (8 pts) A rectangular storage container with an open top is to have a
volume of 10 ft3 . The length of its base is twice the width. Material for
the base costs $9 per square ft. Material for the sides costs $6 per square
ft. Find the cost of materials for the cheapest such container. Round your
answer to the nearest cent.
Let x be the width and y be the height of the box. Then the volume is
V = x · 2x · y = 2x2 y = 10.
Therefore
10
5
=
.
2x2
x2
On the other hand, the area of the base is 2x2 and the total area of sides
is xy + 2xy + xy + 2xy = 6xy. Therefore the cost for the material is
180
5
.
C = 9 · 2x2 + 6 · 6xy = 18x2 + 36xy = 18x2 + 36x · 2 = 18x2 +
x
x
y=
Note that at a minimum, C has a critical point.
dC
= 36x − 180x−2 .
dx
dC
= 0 ⇔ 36x − 180x−2 = 0 ⇔ 36x = 180x−2 ⇔ 36x3 = 180
dx
√
180
3
⇒ x3 =
= 5 ⇒ x = 5.
36
Now the minimum cost is
√
3
C( 5) ≈ 157.90.
answer: $157.90
• Describing the volume 2x2 y = 10 by using width and height: 2 pts.
180
• Finding the cost function C = 18x2 +
: 4 pts.
x
dC
• Computing its derivative
= 36x − 180x−2 : 6 pts.
dx
√
3
• Getting a critical number 5: 7 pts.
• Writing the answer $157.90 with an appropriate unit: 8 pts.
5
MATH 1109 R13 Final
Spring 2014 - Moon
(5) Let f (x) = 9x − 6x2 + x3 .
(a) (3 pts) Find all critical numbers of f .
f 0 (x) = 9 − 12x + 3x2 = 3(x2 − 4x + 3) = 3(x − 1)(x − 3)
f 0 (x) = 0 ⇔ x = 1, x = 3
• Computing the derivative f 0 (x) = 9 − 12x + 3x2 : 1 pt.
• Finding all critical numbers x = 1, 3: 3 pts.
(b) (2 pts) Find the intervals of increase/decrease.
x
x<1 1 1<x<3 3 x>3
0
f (x)
+
−
+
Therefore f is increasing on (−∞, 1) and (3, ∞), and it is decreasing
on (1, 3).
(c) (2 pts) Find all local maxima/minima.
f 00 (x) = −12 + 6x
f 00 (1) = −6 < 0, f 00 (3) = 6 > 0
Therefore f (1) = 4 is a local maximum and f (3) = 0 is a local minimum.
(d) (3 pts) Identify the intervals of concavity.
f 00 (x) = −12 + 6x = 0 ⇔ x = 2
x<2 2 x>2
x
f (x)
−
+
So f is concave upward on (2, ∞)and concave downward on (−∞, 2).
• Computing the second derivative f 00 (x) = −12 + 6x: 1 pt.
• Determining the intervals of concavity: 3 pts.
00
6
MATH 1109 R13 Final
Spring 2014 - Moon
(e) (4 pts) Sketch the graph of y = f (x).
• Plotting precise local maximum, minimum, and inflection points:
2 pts.
• Sketching a curve with appropriate concavity: 4 pts.
7
MATH 1109 R13 Final
Spring 2014 - Moon
(6) (7 pts) Use the right point rule (with 5 intervals) to estimate the area
under the graph y = 2x2 from 1 to 3.
If we divide the interval [1, 3] into 5 subintervals, then each interval
has length 2/5 = 0.4. So the right endpoint of the intervals are
1.4, 1.8, 2.2, 2.6, 3.
On each point, the value of x2 + x is given by the following table:
x 1.4 1.8 2.2
2.6
3
y 3.92 6.48 9.68 13.52 18
Therefore the estimation of the area is
area ≈ 3.92 · 0.4 + 6.48 · 0.4 + 9.68 · 0.4 + 13.52 · 0.4 + 18 · 0.4 = 20.64
•
•
•
•
Finding the length of small subinterval 0.4: 1 pt.
Computing the endpoints of subintervals 1.4, 1.8, 2.2, 2.6, 3: 3 pts.
Calculating the value of y at each right endpoint: 5 pts.
Estimating the area and getting 20.64: 7 pts.
(7) (a) (5 pts) Find the definite integral
Z 2
x2 − 2x + 4 dx.
1
Find the precise value, or round the answer to three decimal places.
Z
2
i2
1 3
2
x − 2x + 4 dx = x − x + 4x
3
1
1
1 3
1
10
=
2 − 22 + 4 · 2 −
−1+4 =
≈ 3.333
3
3
3
2
1
• Finding the antiderivative x3 − x2 + 4x: 3 pts.
3
10
• Getting the answer : 5 pts.
3
8
MATH 1109 R13 Final
Spring 2014 - Moon
(b) (5 pts) Find the indefinite integral.
Z
2
xe3x dx
du
dx = 6x dx
dx
Z
Z
Z
1
u
u
dx = xe dx = e x dx = eu du
6
1 3x2
1 u
= e +C = e +C
6
6
u = 3x2 ⇒ du =
Z
xe3x
2
• Finding an appropriate substitution u = 3x2 : 1 pt.
• Computing du = 6x dx: 2 pts.
1
• By applying substitution rule, obtaining eu + C: 4 pts.
6
1 3x2
• Getting the answer e + C: 5 pts.
6
(c) (5 pts) Find the definite integral
Z 2
x2 (x3 + 2)4 dx.
0
u = x3 + 2 ⇒ du = 3x2 dx
Z
2
3
4
Z
x (x + 2) dx =
Z
2 4
x u dx =
Z
u4 x2 dx
1 1
1
1
du = · u5 + C = (x3 + 2)5 + C
3
3 5
15
Z 2
i2
1
1
1
x2 (x3 + 2)4 dx = (x3 + 2)5 = (23 + 2)5 − (03 + 2)5
15
15
15
0
0
99968
=
≈ 6664.53
15
=
u4
• Finding an appropriate substitution u = x3 + 2: 1 pt.
• By using substitution rule, obtaining the antiderivative
2)5 + C: 4 pts.
• Getting the answer
99968
: 5 pts.
15
9
1 3
(x +
15
MATH 1109 R13 Final
Spring 2014 - Moon
(8) A bicycle is moving along a straight line so that its velocity at time t
seconds is v(t) = 6t2 − 12t ft/s.
(a) (3 pts) Find the displacement of the bicycle during the time period
1 ≤ t ≤ 5.
Z 5
Z 5
i5
v(t) dt =
6t2 − 12t dt = 2t3 − 6t2
displacement =
1
1
1
= (2 · 53 − 6 · 52 ) − (2 · 13 − 6 · 12 ) = 104 ft
5
Z
• Writing the definition
v(t) dt of the displacement: 1 pt.
1
• Finding the antiderivative 2t3 − 6t2 : 2 pts.
• Getting the answer 104 ft and writing it with an appropriate
unit: 3 pts.
(b) (2 pts) On the same time period, find the average velocity.
Z 5
1
1
average =
v(t) dt = · 104 = 26 ft/s
5−1 1
4
• Writing the answer 26 ft/s with an appropriate unit: 2 pts.
(c) (5 pts) Find the distance traveled during the same time period.
v(t) = 6t2 − 12t = 6t(t − 2) = 0 ⇔ t = 0, 2
0 < t < 2 ⇒ v(t) < 0, t > 2 ⇒ v(t) > 0
Z
5
Z
|v(t)| dt =
moving distance =
2
Z
−v(t) dt +
=
1
5
|v(t)| dt
1
Z
2
2
Z
2
−6t + 12t dt +
v(t) dt =
2
5
Z
|v(t)| dt +
1
Z
2
1
5
6t2 − 12t dt
2
i2 i5 3
2
3
2
= −2t + 6t
+ 2t − 6t
2
1
3
2
3
2
= (−2 · 2 + 6 · 2 ) − (−2 + 6) + (2 · 5 − 6 · 5 ) − (2 · 23 − 6 · 22 ) = 112 ft
• Determining the sign of v(t): 1 pt.
Z
• Writing the formula for the moving distance
5
|v(t)| dt: 2 pts.
1
• Finding the integral without absolute value sign for the moving distance: 4 pts.
• Getting the answer 112 ft and writing it with an appropriate
unit: 5 pts.
10
MATH 1109 R13 Final
Spring 2014 - Moon
(9) A retiree is paid $1000 per month by an annuity for 30 years.
(a) (6 pts) If the income is invested in an account that earns 5% interested compounded continuously, what is the future value of the income after 30 years? Round the answer to the nearest cent.
The income stream function is f (t) = 1000 × 12 = 12000.
Z 30
0.05·30
12000e−0.05t dt
FV = e
0
1.5
12000
=e
i30 1
−0.05t
e
−0.05
0
1
1
−1.5
0
1.5
12000
e
− 12000
e = 835605.38
=e
−0.05
−0.05
future value: $835605.38
• Finding the income stream function f (t) = 12000: 1 pt.
Z 30
0.05·30
• Setting the future value formula F V = e
12000e−0.05t dt:
4 pts.
• Getting the answer $835605.38: 6 pts.
0
(b) (4 pts) The retiree may choose another option, which is a one-time
withdrawal of $190,000. Which option is better?
The present value of the annuity is
Z 30
FV
PV =
12000e−0.05t dt = 1.5 = 186448.76.
e
0
Because the present value of the annuity is smaller than $190,000,
we should choose the one-time withdrawal.
Z
30
12000e−0.05t dt: 1 pt.
• Setting the present value formula
0
• Evaluating the present value $186448.76: 3 pts.
• Making a correct decision: 4 pts.
• Alternatively, you may compute the future value of the $190,000
and compare it with the previous future value.
11