SAMPLE EXAM #2
Business Calculus
SHOW ALL WORK – NO NEED TO SIMPLIFY ANSWERS
√
1. (20 points) Using differentials, approximate 4.1.
√
1
Solution: Let f (x) = x, let x = 4 and let ∆x = 4.1 − 4 = 10
. Notice that
1
f 0 (x) = √ and, using differentials,
2 x
√
√
1
1
81
0
4.1 = f (4.1) ≈ f (4) + f (4)∆x = 4 + √
= .
| {z }
40
2 4 10
df
2. (20 points) Find [xx ]0 .
Solution:
h
i0 x ln(x) 0
1
x 0
ln(x) x
x ln(x)
[x ] = e
= e
= xx (ln(x) + 1) .
=e
1 ln(x) + x
x
0
3. (20 points) Find log10 (x2 + 1) .
Solution:
0
ln(x2 + 1)
ln((10)
0
1 =
ln(z 2 + 1)
ln(10)
1
1
2x
=
2x = 2
.
2
ln(10) x + 1
(x + 1) ln(10)
0
log10 (x2 + 1) =
Business Calculus
Sample Exam #2
4. (20 points) A stamp collection is worth $1,200 and its value increases linearly
at $200 a year. If the prevailing interest rate remains constant at 8% per year,
compounded continuously, when will it be most advantageous to sell the stamp
collection? (Section 4.4, Problem 50)
Solution:
Value
Present Value
0
0
=
=
=
=
1, 200 + 200t
P (t) = (1.200 + 200t)e−0.08t
P 0 (t) = 200e−0.08t + (1, 200 + 200t)e−0.08t (−0.08)
200 − 96 − 16t
13
104
t =
= .
16
2
Evaluate the following:
Z 9
5. (20 points)
x3/2 dx.
1
Solution:
Z
9
x
3/2
1
Z
6. (20 points)
√
2
dx = x5/2
5
1
484
2
.
= (243 − 1) =
5
5
x2
dx.
x3 − 3
Solution: Let u = x3 − 3. Then
Z
9
x2 du
1
√
=
2
3
u 3x
Z
du
dx
u−1/2 du =
= 3x2 and dx =
du
. Substituting,
3x2
1
2√ 3
2u1/2 + C =
x − 3 + C.
3
3
Business Calculus
Sample Exam #2
7. (20 points) What are the dimensions of the largest box (in volume) one can
make with $100 if each square foot of top or bottom cost $2, if each square foot
of side cost $1, and if the width equals the length?
Solution: Want: x and y where x = length = width and y = height.
Know:
V = volume = x2 y
to be maxed
C = cost
= top + bottom + 4sides
= 2x2 + 2x2 + 4xy
the restraint
To Do: From the constraint one has
y=
100 − 4x2
25 − x2
=
4x
x
Thus
V (x) = x
2
25 − x2
x
= 25x − x3 .
Maximizing V (x) for 0 ≤ x ≤ 5 one sees
0
0 = V (x) = 25 − 3x
2
⇒
√
5 3
5
.
x= √ =
3
3
Since the max volume is not at 0 or
5 and since √V 0 (x) exists everywhere, the
√
max volume dimensions are x = 5 3 3 and y = 103 3 .
8. (20 points) If you throw a rock upward and one second later its velocity is 16
feet per second (upward), and two seconds later it is 32 feet high. How high will
it go? Remember, acceleration due to gravity is -32 feet per second squared.
Business Calculus
Sample Exam #2
Solution:
Z
v(t) =
Z
−32 dt = −32t + C
a(t) dt =
16 = v(1) = −32 + C ⇒ C = 48
v(t) = −32t + 48
Z
Z
p(t) =
v(t) dt = −32t + 48 dt = −16t2 + 48t + K
32 = p(2) = −64 + 96
p(t) = −16t2 + 48t.
⇒
K=0
Let t0 be when rock is at it highest. Then 0 = v(t0 ) = −32t0 +48 or t0 = 3/2.
Thus, at its highest point the rock is p(3/2) = −16(3/2)2 + 48(3/2) = 36 feet
high.
9. (20 points) What is the present value of an annuity that pays at a rate of $10,000
a year for ten years worth if the prevailing interest rate remains constant at 4%
a year, compounded continuously? (Section 5.5, Problem 30)
Solution:
Z
Present Value =
10
10, 000e−0.04t dt
0
= −10, 000 25e−t/25
−2/5
= −250, 000(e
2
10
0
− 1) = 82, 420.
10. (20 points) Assume D(x) = 65−x2 and S(x) = x3 +2x+5. Find the equilibrium
price, pe and the consumers’ surplus. (Section 5.5, Problem 16)
Business Calculus
Sample Exam #2
Solution:
65 − x2e = D(xe ) = S(xe ) =
x2e
+ 2xe + 5
3
4 2
x + 2xe − 60 = 0
3 e
4x2e + 6xe − 180 = 0
2x2e + 3xe − 90 = 0
−3±
√
32 −4(2)(−90)
Using the quadratic formula one has xe =
= −3±27
, i.e., xe
2(2)
2(2)
15
equals − 2 or 6. Since xe is positive, xe = 6 and pe = D(6) = S(6) = 29.
Furthermore
Z xe
D(x) − pe dx
Consumers’s Surplus =
0
6
Z 6
1 3
2
(65 − x ) − 29 dx = 36x − x
=
= 144.
3
0
0