Download TEST 3

REVIEW
TEST 3
1
Let’s begin Next Slidewith a few
practice derivatives.
To skip practice derivative and go
to implicit derivatives click here.
Please click “ON THE LETTER”
of the correct answer!
2
1. Find the derivative for
y = 3x 2 + 5x - 7
A. y’ = 3x + 5
C. y’ = 6x
C. y’ = 6x + 5
D. y’ = 6x + 5 - 7
E. None of the above
3
No that answer is incorrect.
You need to use the power rule on each of the
terms of the equation.
The power rule is –
If y = x n , then y’ = n x n – 1 .
Please click here to try again.
4
Yes, the answer is y’ = 6x + 5
Good work on using the power rule!
If y = x n , then y’ = n x n – 1 .
Please click here for the next question.
5
2. Find the derivative for
y = 2x – 1 – x – 3
A. y’ = - 2x 0 + 3x – 2
B. y’ = -2x – 2 – 3x – 4
C. y’ = 2x – 2 + 3x – 4
D. y’ = - 2x – 2 + 3x – 4
E. None of the above
6
No that answer is incorrect.
You need to use the power rule on each of the
terms of the equation.
The power rule is –
If y = x n , then y’ = n x n – 1 .
Watch your negative signs!
Please click here to try again.
7
Yes, the answer is y’ = - 2x – 2 + 3x – 4
Good work on using the power rule!
If y = x n , then y’ = n x n – 1 .
Please click here for the next question.
8
3. Find the derivative for
y = (2x 2 – x + 1)(x – 2)
A. y’ = (4x – 1)(1)
B. y’ = 6x 2 - 10x + 3
C. y’ = (2x 2 – x + 1)(1) + (x – 2)(4x – 1)
D. y’ = (4x – 1)(x - 1)
E. None of the above
9
Too bad that answer is incorrect.
You need to use the product rule.
The product rule is –
If
f (x) = F (x) • S (x),
Then f ’ (x) = F (x) • S’ (x) + S (x) • F ’(x)
Please click here to try again.
10
Yes, the answer is y’ = 6x 2 - 10x + 3 or
y’ = (2x 2 – x + 1)(1) + (x – 2)(4x – 1)
Good work on using the product rule
If y = f · s , then y’ = f · s’ + s · f’ .
Please click here for the next question.
11
4. Find the derivative for
2
x 3
y
5x  2
2
(
5
x

2
)(
2
x
)

(
x

3
)(
5
)
A. y ' 
(5x  2)2
(5x  2)( 2x)  ( x 2  3)( 5)
B. y ' 
(5x  2)2
2
5x  4x  15
C. y ' 
( 5x  2) 2
D. y’ = (5x + 2)(2x) – (x 2 + 3) (5)
E. None of the above
12
Sorry, that answer is incorrect.
You need to use the quotient rule.
The quotient rule is –
If
Then
T ( x)
y
B ( x)
B ( x)  T ' ( x)  T ( x)  B ' ( x)
y' 
[ B ( x) ] 2
Please click here to try again.
13
Great, the correct answer is
5x  4x  15
y' 
( 5x  2) 2
2
OR
(5x  2)( 2x)  ( x 2  3)( 5)
y' 
(5x  2)2
You used the quotient rule correctly!!
If
T ( x)
y
B ( x)
then
B ( x)  T ' ( x)  T ( x)  B ' ( x)
y' 
[ B ( x) ] 2
Please click here for the next question.
14
5. Find the derivative for
4
y
x
A. y’ = - 2 x 1/2
C. y’ = - 2
x – 3/2
B. y’ = 4 x – 3/2
D. y’ = 4 x 1/2
E. None of the above
15
Not quite, that answer is incorrect.
Change the equation to remove the radical to –
y = 4x – 1/2
and use the power rule.
Please click here to try again.
16
Hey great, the correct answer is y’ = - 2 x – 3/2
Nice work on the power rule and
negative exponents.
Please click here for the next question.
17
6. Find the derivative for
y = (x 2 – 3x + 6) 5
A. y’ = 5 (x 2 – 3x + 6) 4
B. y’ = (x 2 – 3x + 6) 5 (2x – 3)
C. y’ = (2x - 3) 5
D. y’ = 5 (x 2 – 3x + 6) 4 (2x – 3)
E. None of the above
18
Too bad, that answer is incorrect.
You need to use the chain rule with
u = x 2 – 3x + 6
d n
n  1 du
u  nu
dx
dx
Please click here to try again.
19
OK, the correct answer is
y’ = 5 (x 2 – 3x + 6) 4 (2x – 3)
Good use of the chain rule with
u = x 2 – 3x + 6 !!
d n
n  1 du
u  nu
dx
dx
Please click here for the next question.
20
7. Find the derivative for
y
x 3
2
A. y’ = x (x 2 + 3) – 1/2
B. y’ = (x 2 + 3) – 1/2
C. y' 
x
x2  3
D. y’ = 1/2 (x 2 + 3) - ½ (2x)
E. None of the above
21
Too bad, that answer is incorrect.
You need to rewrite the equation without a
radical and then use the chain rule
y = (x 2 + 3) ½ and the let u = x 2 + 3
d n
n  1 du
u  nu
dx
dx
Please click here to try again.
22
Terrrrriffffic!!
Rewrite the equation without a radical and then use
the chain rule - y = (x 2 + 3) ½ and the let u = x 2 + 3
Then
y’ = x (x 2 + 3) – ½ OR
y’ = 1/2 (x 2 + 3) - ½ (2x) OR
y' 
x
x 3
2
Please click here for the next question.
23
8. Implicitly differentiate to find y’ if
0 = 3x 2 – y + 4
A. y’ = - 6x
B. y’ = 6x
C. y’ = 6x - y
D. y’ = 6x - x
E. None of the above
24
Sorry that answer is incorrect.
To implicit differentiate remember to take the
derivative of each term of the function with respect
to x.
If 0 = 3x 2 – y + 4 then
d/dx (0) = d/dx (3x 2 ) – d/dx (y) + d/dy (4)
Please click here to try again.
25
Yes, the answer is y’ = 6x.
d/dx (0) = d/dx (3x 2 ) – d/dx (y) + d/dy (4)
0 = 6x - y’ + 0
so
y’ = 6x
Please click here for the next question.
26
9. Implicitly differentiate to find y’ if
0 = x2 + y2 + x + y + 3
A.
C.
2x  1
y' 
2y  1
y '   2x  2y  1
B.
 2x
y' 
2y
D.
 2x  1
y' 
2y  1
E. None of the above
27
No that answer is incorrect.
To implicit differentiate remember to take the
derivative of each term of the function with respect
to x.
If 0 = x 2 + y 2 + x + y + 3 then
d/dx (0) = d/dx (x 2 ) + d/dx (y 2 ) + d/dx (x) +
d/dx (y) + d/dx (3)
Now finish the derivatives!!
Please click here to try again.
28
Yes indeeeedy, the answer is
 2x  1
y' 
2y  1
If 0 = x 2 + y 2 + x + y + 3 then
d/dx (0) = d/dx (x 2 ) + d/dx (y 2 ) + d/dx (x) +
d/dx (y) + d/dx (3)
0 = 2 x + 2 y y’ + 1 + y’ + 0 and solving for
y’ yields
 2x  1
y' 
2y  1
Please click here for the next question.
29
10. Implicitly differentiate to find y’ if
x ln y + 2y = 2x 3
Please click on the letter and not the equation.
6x  ln y
A. y ' 
x 2
y
6x  ln y
B. y ' 
x2
6x 2  ln y
C. y ' 
y2
6x 2  ln y
D. y ' 
x  2y
2
2
E. None of the above
30
Too bad that answer is incorrect. This one is difficult!!
To implicit differentiate remember to take the
derivative of each term of the function with respect
to x. This problem also involves the product rule!!!
If x ln y + 2y = 2x 3 then
d/dx (x ln y) + d/dx (2y ) = d/dx (2x 3 )
You need the
product rule here.
Now finish the derivatives!!
Please click here to try again.
31
6x  ln y
y' 
xy2
2
Yes, the answer is
If
x ln y + 2y = 2x 3 then
d/dx (x ln y) + d/dx (2y ) = d/dx (2x 3 )
x (1/y) y’ + ln y (1) + 2 y’ = 6x 2 and solving for y’
yields
2
6x  ln y
y' 
xy2
Please click here for the next question.
32
11. How much would a $2,000 investment be
worth in 12 years at 9% compounded
continuously?
A. $5,865.67
B. $5,889.36
C. $5,885.67
D. $5,885.36
E. None of the above
33
Sorry, that answer is incorrect.
Remember that the continuous compound
interest formula is –
A  Pe
rt
And you have all the information except for A.
Please click here to try again.
34
Great, the correct answer is $5,889.36, since
A  P e  2000 e
rt
12 0.09
 $5,889.36
Please click here for the next question.
35
12. How much would you need to invest at 11%
compounded continuously for 8 years to end up
with $5,000?
A. $2,082.25
B. $2,082.91
C. $2,073.91
D. $2,073.25
E. None of the above
36
Not quite, that answer is incorrect.
Remember that the continuous compound
interest formula is –
A  Pe
rt
And you have all the information except for P.
Please click here to try again.
37
Hey great, the correct answer is $2,073.91, since
A  P e or P  A
rt
P  5000
e
0.11 8
 5000
e
rt
2.4109
so
 $2,073.91
Please click here for the next question.
38
13. What continuously compounded interest
rate is needed to double your money in 13
years?
A. 5.54%
B. 2.32%
C. 5.25%
D. 5.33%
E. None of the above
39
Too bad, that answer is incorrect.
Remember that the continuous compound
interest formula is –
A  Pe
rt
And you have all the information except for r.
This will be a solution that needs natural logs.
Please click here to try again.
40
OK, the correct answer is 5.33%, since
 
A
rt
A  P e or
 e and ln A  r t and
P
P
ln A
ln 2
P
and r 

 0.0533
t
13
rt
 
 P  r
ln A
t
Remember your money
is doubling so A/P = 2
Please click here for the next question.
41
14. An automobile depreciates at a rate of 20%
per year. How long will it take for the used value
to be one-half the new value?
A. 3.1 years
B. 2 years
C. 0.9 years
D. 2.9 years
E. None of the above
42
Not quite, that answer is incorrect.
Remember that the depreciation formula is –
A  P (1  r )
t
And you have all the information except for t.
This will be a solution that needs logs.
Please click here to try again.
43
Terrrrriffffic, the correct answer is 3.1 years, since
 
A
A  P (1  r ) or
 (1  r ) t and ln A  ln (1  r )t
P
P
A
ln
t
P t
A
and ln
 ln (1  r )  t ln (1  r ) and
P
ln (1  r )
t
 
 
 A
ln  
ln 0.5
P

and t 

 3.1
ln (1  r ) ln (0.8)
Please click here for the next question.
44
15. How much would a $2,000 investment be
worth in 12 years at 9% compounded monthly?
A. $5865.67
B. $5,889.36
C. $5,885.67
D. $5,885.36
E. None of the above
45
That answer is incorrect. It is a hard one!
Remember that the compound interest formula
nt
is –
r 

A  P 1  
n

And you have all the information except for A.
Please click here to try again.
46
Terrrrriffffic, the correct answer is $5865.67, since
r 

A  P 1  
n

nt
0.09 

 2000  1 

12 

12 12
 $5,865.67
Please click here for the next question.
47
16. How much would you need to invest at 11%
compounded quarterly for 8 years to end up with
$5,000?
A. $2,073.91
B. $2,075.71
C. $2,098.71
D. $2,198.71
E. None of the above
48
That answer is incorrect.
Remember that the compound interest formula
nt
is –
r 

A  P 1  
n

And you have all the information except for P.
Please click here to try again.
49
Terrrrriffffic, the correct answer is $2,098.71, since
r 

A  P 1  
n

nt
P  5000
so P 
A
r 

1 
n

4 8
0.11 

1 

4 

nt
 $2,098.71
Please click here for the next question.
50
17. Calculate the derivative of
y = 5 ln x
A. y’ = 5x
B. y’ = 5/ln x
C. y’ = 5/x
D. y’ = ln 5x
E. None of the above
51
That answer is incorrect!
Remember that if
y = ln x then y’ = 1/x
Please click here to try again.
52
Grrrrreat, the correct answer is y’ = 5/x
By using the definition of the derivative of the natural
log function.
If y = ln x then y’ = 1/x
The constant 5 is just carried through the problem.
Please click here for the next question.
53
18. Calculate the derivative of
y = 5 ex
A. y’ = 5 e 5 x
B. y’ = 5 e x
C. y’ = 5 / e x
D. y’ = 5 ln x
E. None of the above
54
That answer is incorrect.
Use the definition of the derivative of
the exponential function
If y = e x then
y’ = e x .
Please click here to try again.
55
WOW, the correct answer is indeed y’ = 5 e x.
By using the definition of the derivative of
the exponential function If y = e x then
y’ = e x .
The constant 5 is just carried through the
problem.
Please click here for the next question.
56
19. Calculate the derivative of
y = 2 ln x + 3 e x
A. y’ = 2 ln x + 3 e x
C. y’ = 2/x + 3
ex
B. y’ = 2/x + e x
D. y’ = 2 ln x + e x
E. None of the above
57
That answer is incorrect.
This problem is a combination of the previous two
problems.
Please click here to try again.
58
Great, the correct answer is y’ = 2/x + 3 e x, since the
derivative of 2 ln x is 2/x and the derivative of
3 e x is 3 e x.
Please click here for the next question.
59
20. Calculate the derivative of
ye
A. y '  e
C. y '  e
5 x3
5 x3
 5x
3
 15 x
2
5 x3
B. y '  e
D.
5 x3
y'  e
 5 x2
5 x3
 x2
E. None of the above
60
No that answer is incorrect.
Remember formula for the derivative of e raised
to a functional power If y = e u then y’ = e u du/dx
Please click here to try again.
61
Yes! Yes! Yes!, the correct answer is,
y'  e
5 x3
 15 x2
since
If y = e u then y’ = e u du/dx
Letting u = 5 x 3 gives a chain of du/dx = 15 x 2
and then y’ = e u du/dx
y'  e
5 x3
 15 x2
Please click here for the next question.
62
21. Calculate the derivative of
ye
A. y '  e
C. y '  e
x2  3x
x2  3x
x2  3x
B. y '  e
 2x  3
x2  3x
D. y '  e
x2  3x

 x2  3x
 2x
E. None of the above
63

Sorry that answer is incorrect.
Remember formula for the derivative of e raised
to a functional power If y = e u then y’ = e u du/dx
Please click here to try again.
64
Yes, the correct answer is y '  e
since
x2  3x
 2x  3
If y = e u then y’ = e u du/dx
Letting u = x 2 + 3x gives a chain of du/dx = 2x + 3
and then y’ = e u du/dx
y'  e
x2  3x
 2x  3
Please click here for the next question.
65
22. Calculate the derivative of
y = ln x 5
A. y’ = 5x
B. y’ = 1 / (x 5)
C. y’ = 1 / ( 5x 4 )
D. y’ = 5 / x
E. None of the above
66
Too bad that answer is incorrect.
Remember formula for the derivative of the
natural log function If y = ln u then y’ = 1 / u (du/dx)
Please click here to try again.
67
OK, the correct answer y’ = 5 / x.
I used the formula for finding the derivative of the ln u.
If y = ln u then y’ = 1 / u (du/dx)
Let u = x 5 and then du/dx = 5 x 4 .
If y = ln x 5 then y’ = 1/ ( x 5 )  (5 x 4 ) = 5 / x
An interesting aside. You can rewrite this problem
as follows and make it easier to do.
y = ln x 5 = 5 ln x.
Please click here for the next question.
68
23. Calculate the derivative of
y = ln (x 2 – 4)
2x
A. y '  2
x 4
B. y’ = 2x ln (2x – 4)
C. y’ = 2x ln (x 2 – 4)
D. y’ = 2x ln 2x
E. None of the above
69
No that answer is incorrect.
Remember formula for the derivative of the
natural log function If y = ln u then y’ = 1 / u (du/dx)
Please click here to try again.
70
2x
Correctamundo, the answer is y '  2
x  4
If y = ln u then y’ = 1 / u (du/dx)
Let u = x 2 - 4 and then du/dx = 2x .
If y = ln (x 2 – 4) then y’ = 1/(x 2 - 4)  2x = 2x/ (x 2 - 4)
Please click here for the next question.
71
24. Calculate the derivative of
y = x e – 2x
A. y’ = e – 2x
B. y’ = e – 2x ( 1 – 2x)
C. y’ = x e – 2x ( - 2) + e – 2x
D. y’ = x e – 2x + e – 2x
E. None of the above
72
No that answer is incorrect.
You will need to use the product rule and the chain
rule on this one. Be careful.
If y = f · s then y’ = f · s’ + s · f’ AND the second
function y = e – 2x requires the chain rule!!!
Please click here to try again.
73
Yes, the answer is y’ = e – 2x ( 1 – 2x) or
y’ = x e – 2x ( - 2) + e – 2x
We used the product rule on this one:
Let
f=x
and
s = e – 2x then
f’ = 1
and
s’ = e – 2x (-2) and
y’ = x e – 2x ( - 2) + e – 2x = e – 2x ( 1 – 2x)
Please click here for the next question.
74
25. Calculate the derivative of
y = e – x ln x
A. y’ = e – x ln x
C. y’ =
e – x/x
-
e–x
B. y’ = e – x - e – x ln x
D. y’ = e – x/x - e – x ln x
E. None of the above
75
No that answer is incorrect.
You will need to use the product rule and the chain
rule on this one. Be careful.
If y = f · s then y’ = f · s’ + s · f’ AND the second
function y = e – 2x requires the chain rule!!!
Please click here to try again.
76
Yes, the answer is y’ = e – x/x - e – x ln x.
We used the product rule on this one:
Let
f = e–x
and
s = ln x
then
f’ = e – x (- 1)
and
s’ = 1/x
and
y’ = e – x/x - e – x ln x.
Please click here for the next question.
77
26. Calculate the derivative of
ln x  1
y
x
e
A.
e x  (ln x  1)
y' 
(e x ) 2
e x  e (ln x  1)
C. y' 
(e x )2
x
x
1 x  (ln x  1)
B. y' 
(e x )
e x x  e x (ln x  e x )
D. y' 
(e x )2
E. None of the above
78
Too bad that answer is incorrect.
You will need to use the quotient rule on this one.
If y = t / b then
b t '  t b '
y' 
b2
Please click here to try again.
79
e x x  e x (ln x  1)
Yes, the answer is y' 
(e x )2
1 x  (ln x  1)
y' 
(e x )
Using the quotient rule –
b = ex
and
t = ln x + 1
b’ = e x
and
t’ = 1/x
or
so
e x x  e x (ln x  1)
y' 
(e x )2
Please click here for the next question.
80
APPLICATIONS
Now, let’s try some application problems
81
27. A drug is injected into the bloodstream. The
concentration of the drug after x hours is
given by
0.20x
C( x) 
x 4
2
for 0  x  10.
A. Find the marginal concentration after 3
hours (4 decimal places).
A. C’ (x) = 0.0462 m/cm 3
B. C’ (x) = – 0.1746 m/cm 3
C. C’ (x) = – 0.0059 m/cm 3
D. C’ (x) = 0.5428 m/cm 3
E. None of the above
82
Sorry that answer is incorrect.
To find the marginal concentration graph the given
function and under the CALC menu use the dy/dx
choice at x = 3.
Please click here to try again.
83
Yes, the answer is C’ (x) = – 0.0059 m/cm 3
from your graphing calculator.
NOTE: You will need this answer for the next
question.
Please click here for the next question.
84
28. A drug is injected into the bloodstream. The
concentration of the drug after x hours is
given by
0.20x
C( x) 
x 4
2
for 0  x  10.
B. Interpret the results of the previous
question.
Take a minute and try to write a response.
Go to a representative correct answer.
85
The drug concentration after three hours is decreasing
by about 0.0059 m/cm 3 for the next hour.
Please click here for the next question.
86
29. If consumer demand is given by
D (p) = 10000 e – 0.02p , where p is the price in
dollars and d is the demand in units, find the
price that maximizes Consumer
Expenditure.
A. p = $50.00
B. p = $53.11
C. p = $55
D. p = $68.56
E. None of the above
87
Too bad that answer is incorrect.
Consumer expenditure is consumer demand times
the price. Graph consumer expenditure on your
calculator and find the max.
Please click here to try again.
88
Yes, the answer is $58.51
Consumer expenditure is consumer demand times
the price. Graph consumer expenditure on your
calculator and find the max.
CE = 10000 p e – 0.02p
Please click here for the next question.
89
30. Find the relative rate of change of the
function f(t) = 100 e 0.2t when t = 15.
A. P’ (x) = .15 or 15%
B. P’ (x) = .20 or 20%
C. P’ (x) = .25 or 25%
D. P’ (x) = .27 or 27%
E. None of the above
90
No that answer is incorrect.
The relative rate of change is f ’ (t) / f (t). Try that on
your calculator.
Please click here to try again.
91
Yesaroonie, the answer is 0.2 or 20%
The relative rate of change is f ’ (t) / f (t) and using
my calculator Graph f(t)  100 e 0.2t
f ' (15)
401.71074
RRC 

 0.2  20%
f (15)
2008.55369
Please click here for the next question.
92
31. What is the meaning of life?
Take a minute and write what you think is the
meaning.
Go to a representative correct answer.
93
The average profit is decreasing by about 14 people
for the sale of the 51st calculator on any Tuesday when
it is raining.
Please click here for the next question.
94
32. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
A. Find the revenue function.
A. R (x) = x (6000 - x)/30
B. R (x) = (6000 - x)/30x
C. R (x) = x (72000 + 60x)
D. R (x) = p (x) + C (x)
E. None of the above
95
Sorry that answer is incorrect.
To find the revenue function use R (x) = xp, where p
is the price-demand function.
Please click here to try again.
96
You have what it takes.
The correct answer is
R (x) = xp
= x (6000 - x)/30
NOTE: You will need this answer for the next
question.
Please click here for the next question.
97
33. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
A. Find the marginal revenue function.
A. R’ (x) = 60000 – x/30
B. R’ (x) = p’ (x) – C’ (x)
C. R’ (x) = 200 – x/15
D. R’ (x) = 200 – x/30
E. None of the above
98
Not the correct answer.
To find the marginal revenue function you need to
find the derivative of the revenue function from the
previous problem.
R (x) = x (6000 - x)/30
Please click here to try again.
99
Great calculus work.
The correct answer is
the derivative of R (x) = x (6000 - x)/30
R (x) = x (6000 - x)/30
= 200x – x 2/30 and
R’ (x) = 200 – 2x/30
= 200 – x/15
NOTE: You will need this answer for the next
question.
Please click here for the next question.
100
34. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
C. Find the marginal revenue at x = 1500.
A. $25
B. $100
C. $75
D. $125
E. None of the above
101
Sorry incorrect. Try the following hint.
Plug 1500 into the marginal revenue equation
(answer to the previous problem) for x and solve for
R’ (x),
Please click here to try again.
102
Great work.
The marginal revenue equation is
R’ (x) = 200 – x/15 and
R’ (1500) = 200 – 1500/15
= 200 - 100
= $100
NOTE: You will need this answer for the next
question.
Please click here for the next question.
103
35. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
D. Write a brief interpretation of the result
of the previous problem.
Go ahead, I’ll wait for you.
Click here to see a representative correct answer.
104
At a sales rate of 1500, the revenue will be about
$100 on the sale of the next gadget.
Please click here for the next question.
105
36. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
E. Find the profit function.
A. P (x) = 260x – x 2/30 - 72000
B. P (x) = C (x) – R (x)
C. P (x) = 140x – x 2/30 - 72000
D. P (x) = 140x – x/30 - 72000
E. None of the above
106
Too bad, that is incorrect.
The profit function is found by subtracting cost from
revenue, or
P (x) = R (x) – C (x)
Please click here to try again.
107
Well done!
The correct answer is P (x) = 140x – x 2/30 - 72000
P (x) = R (x) – C (x)
= (200x – x 2/30) – (72000 + 60x)
= 140x – x 2/30 - 72000
NOTE: You will need this answer for the next
two questions.
Please click here for the next question.
108
37. The price-demand and cost equations for
producing gadgets are given by:
p (x) = (6000 - x)/30 and C (x) = 72000 + 60x
for 0  x  6,000.
E. Find the marginal profit function.
A. P’ (x) = 140x – x 2/30 - 72000
B. P’ (x) = 140 – x/30
C. P’ (x) = 140 – x/15
D. P’ (x) = 140x – x/15
E. None of the above
109
Sorry that is incorrect.
The marginal profit is the derivative of the profit and
the profit was found in the previous problem.
Please click here to try again.
110
Well done!
The correct answer P’ (x) = 140 – x/15
Take the derivative of P in the previous problem.
P (x) = 140x – x 2/30 – 72000 and
P’ (x) = 140 – x /15
Please click here for the next question.
111
38. Use the profit equation from #36 to find
when the maximum profit occurs.
A. $ 2,100
B. 58.9
C. $75,000
D. 2100
E. None of the above
112
No that is incorrect.
The profit equation from #36 is below. Find where
the profit is a maximum
P (x) = 140x – x 2/30 - 72000
Please click here to try again.
113
Sorry that is not the correct answer, but it is close.
You were asked to find where the maximum profit
occurs. You want the x value not the x value.
Please click here to try again.
114
Terrific!
The correct answer 2100.
Did you use your calculator to
find that?
You will probably be asked to
find the price that gives the
max profit. Plug 2100 into the
price equation in problem #36.
0 < x < 100 and 0 < y < 200.
Please click here for the next question.
115
39. Given the price-demand equation
f (p) = 12,000 – 10 p 2.
A. Find the elasticity of demand.
A.
C.
20 p 2
12000  10p 2
20 p
12000  10p 2
2 p2
B.
1200  p 2
D.
20 p 2

12000  10p 2
E. None of the above
116
Sorry that answer is incorrect.
The Elasticity of Demand is
p f ' (p )
E (p )  
f (p )
Please click here to try again.
117
You are a derivating genius. The correct answer is
20 p 2
E (p ) 
12000  10p 2
p f ' (p )
p (  20p)
20 p 2
E (p)  


2
2
f (p )
(12000  10p ) 12000  10p
You will need this answer for the next question.
Please click here for the next question.
118
40. Given the price-demand equation
f (p) = 12,000 – 10 p 2.
B. If p = 10, is the demand elastic, inelastic,
or does it have unit elasticity?
A. Elastic
B. Inelastic
D. Unit Elasticity
119
Not the correct answer.
Use the results for the last question and
substitute 10 for p.
Please click here to try again.
120
Great work.
The correct answer is inelastic!
E (10) 
20 10
2
12000  1010
2
2000

 0.18
12000  1000
Since E(p) < 1 the demand is inelastic.
Please click here for the next question.
121
41. A rock is thrown into a pond and causes a
circular ripple. If the radius of the ripple is
increasing at 2 feet per second, how fast is
the circumference changing when the radius
is 10 feet? (Note: C = 2πR.)
A. 25
B. 12.57
C. 6.28
D. 333.33
E. None of the above
122
Sorry incorrect. Try the following hint.
Implicit differentiation the C equation with respect to
t (time) and substitute what you know and solve for
what you don’t know.
Please click here to try again.
123
Great implicit differentiation.
The correct answer is 12.57.
If C = 2r then dC/dt = 2  dr/dt and
dC/dt = 2  2
= 4   12.57
Please click here for the next question.
124
42. A company making calculators has the
following functions: C (x) = 72,000 + 60x
and R (x) = 200x – x 2/30.
C. If production is increasing at 500
calculators per week find the rate of increase
in the cost over time.
A. $25,00
B. $468.75
C. $12,500
D. $ 30,000
E. None of the above
125
Sorry that answer is wrong.
Implicit differentiation the C(x) equation with
respect to t (time) and substitute what you know and
solve for what you don’t know.
Please click here to try again.
126
WOW, yes a tough one!
The correct answer is $30,000.
If C (x) = 72,000 + 60x then dC\dt = 60 dx/dt
And since dx/dt = 500 then dC\dt = (60) (500)
= $ 30,000
Please click here for the next question.
127
43. The price demand equation is
x 2 + 2xp + 25 p 2 = 74,500.
A. If the price is increasing at a rate of $2
per month when the price is $30, find the
rate of change of the demand.
A. 200
B. 175
C. - 8.26
D. - 255
E. None of the above
128
Too bad, that is incorrect AND this is a tuff one!
Implicit differentiation x 2 + 2xp + 25 p 2 = 74,500
with respect to t (time) and substitute what you know
and solve for what you don’t know.
Please click here to try again.
129
Well done!
The correct answer is
x + 2 + 2000/x
If x 2 + 2xp + 25 p 2 = 74,500 [NOTE: when p = 30
x = 200. Use you calculator to confirm that.]
Then 2x dx\dt + 2x dp/dt + 2p dx/dt + 50p dp/dt = 0
And solving for dx/dt (what we want to find) gives
dx

dt
 2x dp
 50 p dp
dt
2x  2p
dt   2 200( 2)  50 ( 30) ( 2)   8.26
2 200  2 ( 30)
Please click here for the next question.
130
44. A new employee is able to assemble N units
after t days according to the relationship:
N (t) = 10(1 – e – 0.5 t )
B. What is the rate of assembly after 5 days?
A. 37
B. 41
C. 0.41
D. 0.37
E. None of the above
131
Sorry that is incorrect.
Remember the derivative is the rate of change.
You want the derivative of the given function
evaluated when x = 5.
Please click here to try again.
132
Well done!
The correct answer 0.41
Did you use your
calculator to find that?
Graph the given
function and use dy/dx
under the “Calc” menu
at x = 5.
Please click here for the next question.
133
That was a lot of review. I hope you found it
helpful. Good luck on the test!
134