MATH 1100: QUANTITATIVE ANALYSIS
FALL 2001 FINAL EXAM
1. Completely calculate the reduced form, and then find all solutions to
x + 2y − z = 1
3x + y = 7
2x − y + z = 2.
Solution:

1
3
2
↓

1
0
0
↓

1
0
0

2 −1 1
1
0 7
−1 1 2
2 −1
−5 3
−5 3

1
4
0

2 −1 1
−5 3
4
0
0 −4
There are no solutions, because the last equation now reads 0 = −4. One
more step gives:


1 0 1/5 0
0 1 −3/5 0 .
0 0
0
1
2. Suppose that a business has continuous income stream with an annual
rate of flow at time t given by f (t) = 100. The interest rate is r = 10%,
compounded continuously. Find the capital value of the income stream.
Solution:
Z ∞
CV =
100e−0.1t dt = $1000.
0
3. Find the quantities x and y of two products which you should buy to
maximize utility U = x2 y 2 subject to all three of the budget constraints
x≥0
y≥0
x + 2y ≤ 20.
Date: April 18, 2003.
1
2
MATH 1100: QUANTITATIVE ANALYSIS
FALL 2001 FINAL EXAM
Hint: these constraints form a triangle. You have to check for critical points
of U inside the triangle, using the first derivatives, check for critical points
on the sides of the triangle using Lagrange multipliers (or perhaps just
common sense), and also check the corners of the triangle. Make sure your
answer is a point (x, y) satisfying all of the budget constraints.
Solution: Differentiating:
∂U
= 2xy 2
∂x
∂U
= 2x2 y.
∂y
For these both to vanish, we need x = y = 0. But there U = 0, which is
clearly a minimum, since x2 ≥ 0 and y 2 ≥ 0 forces U ≥ 0. Similarly on
the sides x = 0 and y = 0 of the triangle, U = 0, so a minimum. On the
remaining side, x + 2y = 20, we can try Lagrange multipliers:
F (x, y, λ) = U + λ(x + 2y − 20).
The partial derivatives are
∂F
= 2xy 2 + λ
∂x
∂F
= 2x2 y + 2λ
∂y
∂F
= x + 2y − 20.
∂λ
These have to vanishing, giving
λ = −2xy 2 = −x2 y.
If x and y don’t vanish, this gives 2y = x. Now we have two linear equations:
2y = x and x + 2y − 20 = 0. The only solution is x = 10, y = 5. This point
solves all three constraints. Now consider the three corners of the triangle:
they all lie on either x = 0 or y = 0 and so have U = 0, a minimum.
Therefore the maximum is U = 2500 at (x, y) = (10, 5).
4. Suppose that the joint cost function of two products is C(x, y) =
Find the marginal cost with respect to each of x and y.
Solution:
∂C
1
=
∂x
(x + 1)(y + 1)
∂C
ln(x + 1)
=−
.
∂y
(y + 1)2
5. Show that
(
3e−3t
f (t) =
0
is a probability density function.
if t ≥ 0
otherwise
ln(x+1)
y+1 .
MATH 1100: QUANTITATIVE ANALYSIS
Solution:
anywhere.
FALL 2001 FINAL EXAM
3
The exponential function is positive, so f (t) is not negative
Z
∞
Z
f (t) dt = 3
−∞
∞
e−3t dt
0
t=∞
e−3t −3 t=0
1
=0−3
3
= 1.
=3
6. Using n rectangles of width 1/n whose left shoulder touches the graph of
f (x) = 2x + 1, approximate the area under the graph of f (x) between
x = 0 and x = 1. DO NOT use any other method to solve this problem:
use rectangles. You might need to know that
n
X
k=1
n
X
1=n
k = n(n + 1)/2.
k=1
Solution: The first triangle has height 1. The second has height 1 + 2/n.
The third has height 1 + 4/n. The k-th rectangle has height f ((k − 1)/n) =
1 + 2(k − 1)/n. The area of the k-th rectangle is
2(k − 1)
base · height = (1/n) · 1 +
n
1
2(k − 1)
= +
.
n
n2
4
MATH 1100: QUANTITATIVE ANALYSIS
FALL 2001 FINAL EXAM
Adding together these areas gives
n X
1
2(k − 1)
total area =
+
n
n2
k=1
n X
1
k
2
=
+2 2 − 2
n
n
n
k=1
=
n
n
n
1X
2 X
2 X
1+ 2
k− 2
1
n
n
n
k=1
k=1
k=1
2 n(n + 1)
2
n
= + 2
− 2n
n n
2
n
n(n + 1) 2n
=1+
− 2
n2
n
n2 + n − 2n
=1+
n2
n2 − n
=1+
n2
1
=2− .
n
7. Suppose that
x + exy = y 2 + ln(y).
Find dy/dx when x = 0 and y = 1.
Solution:
dy
dy
1 dy
1 + exy y + x
= 2y
+
.
dx
dx y dx
Plug in x = 0 and y = 1,
2=3
dy
dx
so
dy
2
= .
dx
3
8. Suppose that $2 is invested at a 10% nominal interest rate, compounded
continuously.
(a) Write an equation describing the value of the investment at t years
since the principal was invested.
(b) Approximate the value of the investment after 30 years, using the
approximation e3 ∼ 20.
(c) How many years does it take for the investment to multiply in value
twenty times over?
(d) How many years does it take for the investment to multiply in value
ninety times over? Use the approximation ln(90) ∼ 4.5.
Solution:
(a)
S = 2e0.10t
MATH 1100: QUANTITATIVE ANALYSIS
FALL 2001 FINAL EXAM
5
(b)
S = 2e3 ∼ 40.
(c) From part (b), 30 years.
(d) 45 years.
9. Suppose that national consumption is $5 billion when disposable income
is zero. Measuring national consumption in billions, suppose that it varies
according to
dC
4y
=
.
dy
C −3
Find the national consumption as a function of disposable income.
Solution: Separating variables we have
(C − 3) dC = 4y dy
which integrates to
(C − 3)2
= 2y 2 + K
2
for K a constant, so
(C − 3)2 = 4y 2 + K
or
C =3±
Plugging in y = 0, we have
p
4y 2 + K.
5=C =3±
√
K
so we have to take the + sign in the ±, and have K = 4. This gives
p
C = 3 + 2 y 2 + 1.