Answers to Review Test 3:
1) Answer in class if someone remembers to ask.
2) a)
f ( x)  x3  9 x 2  15 x
f / ( x)  3x 2  18 x  15  3( x  1)( x  5)  0
x  1, x  5
f(x) is increasing on (  ,1) and (5,  ) and decreasing on (1, 5)
x = 1 gives the relative maximum and x = 5 gives relative minimum
f // ( x)  6 x  18  0 therefore x = 3 is the point of inflection and f(x) is concave down on
(  ,3) and concave up on (3,  ).
f(x)
50
45
40
35
30
25
20
15
10
5
0
-5
-10 0
-15
-20
-25
-30
1
2
3
4
5
6
7
8
3)
f ( x)  x 4  4 x3 You need to consider f(-1) = 5, f(1) = -3, f(0)=0 to determine that the
absolute maximum is at x = -1, and absolute minimum at x = 1.
4) a) Vertical Asymptote at x = -2 since this value gives division by zero.
12 x  12
 4
b) Horizontal asymptote is y = Limit
x 
3x  6
c) This first and second derivative are never equal to zero.
d) Both the first and second derivative are both undefined at x = -2.
e) This first derivative is always negative so the function is always decreasing.
f(x)
10
5
0
-10
-8
-6
-4
-2
0
-5
-10
-15
-20
2
4
6
5) a) Revenue R = 600q – 5q2
Now find the marginal revenue and critical points in order to determine the increasing
and decreasing.
R / (q)  600  10q  0 and q = 60 and revenue is increasing on (0, 60) and decreasing for
all q > 60.
b) The ideal relative maximum occurs at q = 60 but since they are only able to make 50
bikes the realistic maximum will occur at q = 50 since revenue is increasing at that point.
c)
Revenue
20000
15000
10000
5000
0
0
20
40
60
80
100
120
d) Profit = -5q2 + 400q -1500
Take the derivative (marginal profit) and set it equal to zero and solve to get q = 40.
Take the second derivative and verify the q = 40 is indeed the max profit.
e) price = 600 – 5q = 600 – 5(40) = $400.
6) Is omitted
7)
R  25q (0.95) q
R /  25(0.95) q  25q(0.95) q ln(0.95)
 25(0.95) q ( q  ln(0.95))  0
q   ln(0.95)  .05129
8)
2
f ( x)  18 x  x 3
3
/
f ( x)  18  2 x 2  0
x  3
f //  4 x, f // (3)  12, f // (3)  12
Therefore x = -3 is minimum and x = 3 is maximum
9) a) Vertical asymptote is x = 3
b) The first derivative is zero at x = 0 and x = 6.
f(x) increases (, 0) , (6,  ) and decreases on (0, 3), (3,6)
c) Relative max at x = 0 and relative min at x = 6.
d) f(x) is concave down on (  , 3) and concave up on (3, 6)
e) no points of inflection.
Graph:
Graph of f(x)
18
15
12
9
6
3
-6
-5
-4
-3
-2
0
-1 0
-3
1
2
3
4
5
6
7
8
9
10
-6
-9
10) Will be done in class!
11) If production number is increased from q = 105 to 106, then TOTAL revenue is
increased by $128. Use marginal analysis to estimate
R(106) = R(105) + MR(105) = $1128
12) a) Profit = R – C = (408q – 4q2) – (1250 – 8q) = -4q2 +400q -1250
P / =-8q +400 = 0 so q = 50
b) P // =-8 so P is concave down and q = 50 is the maximum situation.
c) price = 408 – 4q = 408 -200 = $208
d) MP(10) = -8(10) + 400 = 320, since MP is positive, Profit is increasing at q = 10
Profit
9000
7000
5000
3000
1000
-1000 0
-3000
20
40
60
80
100
13) F(x) = x 3  12 x  20
a) If x = 0 then y = 20
b) F /  3x 2  12  0
x =  2 F(x) is increasing on ( , 2) , (2,  ) and decreases on (-2,2)
c) Relative maximum is at x = -2 and Relative minimum is at x = 2
d) F //  6 x  0 so x = 0 is the point of inflection and F(x) is concave down on
(, 0) and concave up on (0,  ).
e) Consider the values F(0) = 20, F(2) = 4, F(3) = 12: therefore the absolute
maximum is at x = 0, and absolute minimum is at x = 2.
Graph:
f(x)
-5
-4
-3
-2
50
45
40
35
30
25
20
15
10
5
0
-1 -5 0
-10
-15
-20
1
2
3
4
5
6
7
8
14) Profit = R – C = (1600q – q2)- ( 1500q + 500) = -q2 +100q -500
MP = -2q + 100 = 0 so q = 50. The second derivative is negative so q = 50 is the
ideal maximum profit but since they are only able to make 40 bears the real max profit
will occur at q = 40 and the price they need to charge is
Price = 1600 – 40 = $1560.
15) q = 400 – 4p
a)
p dq
p
4 p
160
2
E

(4) 

  The absolute value is less than 1,
q dp 400  4 p
400  4q 240
3
therefore the demand is elastic at p = 40.
b) To find when the demand is unit elastic set E = -1.
4 p
 1
400  4 p
4 p  400  4 p
400  8 p
p  50
If the price is $50 then you have unit elasticity and maximum revenue.
c) Revenue is R = 400q – 4q2 and MR = 400 – 8q = 0 which gives you the maximum at
q = 50.
16) N, L, N, L, N