ECON 331 - Mathematical Economics - ANSWERS FINAL EXAM
1. (a)- The consumer's utility maximization problem is written as:
max[xy + y 2 + 2x + 2y]
x,y
s.t. 6x + 10y = m, x ≥ 0, y ≥ 0
- The associated Lagrangian function is dened as:
L(x, y, λ) = xy + y 2 + 2x + 2y − λ [6x + 10y − m]
where λ is the Lagrange multiplier.
- The Kuhn-Tucker conditions are written as:
∂L(x, y, λ)
∂L(x, y, λ)
= 0, x ≥ 0,
= y + 2 − 6λ ≤ 0
∂x
∂x
∂L(x, y, λ)
∂L(x, y, λ)
(ii) y
= 0, y ≥ 0,
= x + 2y + 2 − 10λ ≤ 0
∂y
∂y
∂L(x, y, λ)
(iii)
= − [6x + 10y − m] = 0
∂λ
(i)
x
- From (iii): x = m/6 − 5/3y (*).
- Suppose y = 0. From (*) x = m/6 > 0 ⇒ ∂L
= 0 ⇒ λ = 1/3. Plug λ and x in (ii):
∂x
∂L
you nd ∂y = m/6 − 4/3 ≤ 0. However, remember that 0 < m < 40 ⇒ 0 < m/6 − 4/3
which is a contradiction.
- Suppose y > 0 ⇒ ∂L
= 0 ⇒ x + 2y + 2 − 10λ = 0 (**).
∂y
Suppose also x = 0. Then from (*) y = m/10. And from (**) λ = m/50 + 1/5.
Plug y and λ in (i). You nd ∂L
= −m/50 + 4/5 ≤ 0. However, remember that
∂x
8 < m < 40 ⇒ 0 < −m/50 + 4/5 which is a contradiction.
So suppose instead x > 0 ⇒ ∂L
= 0 ⇔ y = 6λ − 2. And from (*) x = m/6 − 10λ + 10/3.
∂x
Plug x and y in (**) and solve for λ. You nd:
x∗ = (40 − m)/24 > 0
y ∗ = (m − 8)/8 > 0
b/c m < 40
b/c m > 8
λ = (m + 8)/48
NB: alternatively, you can directly plug the constraint into the objective function
(say by replacing y by a function of x). You end up with an optimization problem with
one choice variable, x and one non-negativity constraint.
(b) We have seen a result in class that states:
∂U (x∗ , y ∗ )
=λ
∂m
1
We check by calculation that it works here:
µ
¶2
40 − m m − 8
m−8
1
1
2
40 − m m − 8
∗ ∗
U (x , y ) =
×
+
+
+
= m2 + m +
24
8
8
12
4
96
6
3
∗ ∗
∂U (x , y )
1
1
⇒
= m+
(= λ)
∂m
48
6
∂U (x∗ , y ∗ )
7
⇒
=
∂m
12
|m=20
(c) - case where 0 ≤ m ≤ 8: previous solution does not work because it violates the
positivity of x∗ . So we need to study the behavior of U on the boundary.
. x = 0 and y > 0: U (0, y) = y 2 + 2y with y > 0 and 10y = m. So y = m/10 and
U1 = m2 /100 + m/5.
. x > 0 and y = 0: U (x, 0) = 2x with x > 0 and 6x = m. So x = m/6 and U2 = m/3
We can check that U1 < U2 (when m ≤ 8), so the solution is: x∗ = m/6 and y ∗ = 0.
- case where m ≥ 40: similar study needs to be performed as the positivity of y ∗ (found
in b)) is violated:
. same 2 cases and associated solutions! This time, U1 > U2 (when m ≥ 40), hence the
solution is x∗ = 0 and y ∗ = m/10.
2. (a)
In + sM = (In + M )−1 ⇔ (In + sM )(In + M ) = In
⇔ In + M + sM + sM 2 = In
⇔ (1 + s)M + s(3M ) = 0
b/c by assumption M 2 = 3M
⇔ (1 + 4s)M = 0
⇔ 1 + 4s = 0 or M = 0
1
⇔ s = − or M = 0
4
To conclude, for any matrix M s.t. M 2 = 3M , (In − M/4) is the inverse matrix of
(In + M ).
(b) (i)
AXB = XB + C ⇔ (AX − X)B = C
⇔ (A − In )X = CB −1
⇔ X = (A − I)−1 CB −1
(ii)
A−In =
Ã
1
1
1 −1
!
Ã
; (A−In )−1 =
1/2
1/2
1/2 −1/2
2
b/c B is invertible by ass.
b/c A − In is invertible by ass.
!
apply the inverse formula for (2,2)-matrices

and B −1

1 0 0


=  0 2 0  apply the inverse formula for diagonal matrices
0 0 4
Ã
X =
Ã
=
Ã
=
1
1
1 −1
0 2
1 −1
0 4
1 −2

1 0 0
1 1 2


×
× 0 2 0 
−1 3 1
0 0 4


!
1 0 0
3/2


× 0 2 0 
1/2
0 0 4
!
6
2
!
Ã
!

3. - Taylor's approximation: we dene a function and Taylor provides a polynomial
approximation of this function around a point of interest. Here I dene f (x) = x1/3
and I want to approximate it around x0 = 8. The second-order approximation is:
1
f (x) ∼ f (8) + f 0 (8)(x − 8) + f 00 (8)(x − 8)2
2
1 −2/3
1 2 × 8−5/3
f (8.1) ∼ 2 + 8
× (8.1 − 8) − ×
× (8.1 − 8)2
3
2
3×3
1
1
2
= 2+
× 0.1 −
× 0.1
12
144 × 2
= 2 + 0.008333 − 0.000035
= 2.0083
- Newton's algorithm: we dene an equation that has solution at x = 8.11/3 and
Newton's provides an approximate solution. Here I dene g(x) = x3 − 8.1 and the
equation g(x) = 0 has solution at x = 8.11/3 . The algorithm is given by:
g(xi )
(whenever g 0 6= 0)
g 0 (xi )
241
8 − 8.1
=
∼ 2.00833
= 2−
2
3×2
120
241 (241/120)3 − 8.1
=
−
∼ 2.0083
120
3 × (241/120)2
xi+1 = xi −
x1
x2
3
4. (a) - We can build up the following table of revenues for successive years:
Y ear
0
1
2
..
.
n
Revenue
A
A − pA = (1 − p)A
(1 − p)A − p[(1 − p)A] = (1 − p)A[1 − p] = (1 − p)2 A
..
.
(1 − p)n A
So the revenue in year n can be expressed as Rn = (1 − p)n A. (formula can be proved
by induction - not requested here.)
- The company operates as long as the yearly revenue exceeds the yearly cost, ie as long
as (1 − p)n A > K . So we need to nd the largest integer n∗ (representing the largest
number of years) that satises this equation.
(1 − p)n A > K ⇔ ln [(1 − p)n A] > ln(K)
(K > 0!!)
⇔ n ln(1 − p) + ln(A) > ln(K)
⇔ n ln(1 − p) > ln(K) − ln(A)
µ ¶
K
⇔ n ln(1 − p) > ln
A
¡K ¢
ln A
⇔ n<
(0 < (1 − p) < 1, so ln(1 − p) < 0, so inequality changes!!)
ln(1 − p)
So n∗ is the largest integer satisfying the above equation.
(b) The total prot from year 0 to n∗ is simply the dierence between the total revenue
and the total cost from year 0 to n∗ . The total revenue is the sum of the yearly prots
from year 0 to n∗ , ie:
" n∗
#
n∗
n∗
X
X
X
R =
Ri =
[(1 − p)i A] =
(1 − p)i A
i=0
i=0
i=0
∗
1 − (1 − p)n +1
=
A
1 − (1 − p)
∗
1 − (1 − p)n +1
A
=
p
( use the hint with x = 1 − p)
The cost is the same each year, so the total cost is simply: C = (n∗ + 1) × K
Finally, the total prot is:
1 − (1 − p)n
π=
p
∗ +1
A − (n∗ + 1) × K
(c) (i) Plug the given values in the equation found in (a) to get:
¡ ¢
¡ 5000000 ¢
ln K
ln
A
7000000
=
= 16.65
ln(1 − p)
ln(0.98)
4
n∗ is the largest integer that is smaller than 16.65, so n∗ = 16.
(ii) The associated prot is:
π=
1 − (0.98)17
× 7000000 − 17 × 5000000 = 16.74 × 106
0.02
(with n∗ = 14, we get: π = 6.5 × 106 ).
5. (a) The coecient matrix of the system is:


1
3
4


A= 2
2
1 
3 −3 −9
and its determinant is calculated through an expansion in co-factors according to the
rst column (other expansions according to any line or column is acceptable - same
nal answer):
¯
¯
¯
¯
¯
¯
¯ 2
¯ 3
¯ 3 4 ¯
1 ¯¯
4 ¯¯
¯
¯
¯
¯
det(A) = ¯
¯ − 2¯
¯ + 3¯
¯
¯ −3 −9 ¯
¯ −3 −9 ¯
¯ 2 1 ¯
= −18 + 3 − 2(−27 + 12) + 3(3 − 8)
= 0
(b)(i) From (a), the coecients matrix associated with the system has a zero-determinant.
Hence, the system does not have a unique solution: it may have no solution or an innite number of solutions, but we cannot tell yet. It cannot be solved by matrix-inversion
and the Cramer's rule cannot be used.
(ii) We solve the system using the general Gaussian elimination method:


 x + 3y + 4z = b1
2x + 2y + z = b2


3x − 3y − 9z = b3


 x + 3y + 4z = b1
⇔
− 4y − 7z = b2 − 2b1 (L2 − 2L1 )


− 12y − 21z = b3 − 3b1 (L3 − 3L1 )


4z = b1
 x + 3y +
⇔
y + 7/4z = b1 /2 − b2 /4
(−L2 /4)


0 = b3 − 3b1 − 3b2 + 6b1 (L3 − 3L2 )
The system has solutions if and only if the coecients b1 , b2 , and b3 satisfy the last
equation: 3b1 − 3b2 + b3 = 0.
5
6. (a) The domain is dened by couples (x, y) that satisfy 2 equations: (i) x2 + y 2 ≤
1; (ii) x ≥ 0. To sketch the domain, we use the hint! We know that couples (x, y)
satisfying x2 + y 2 = 1 lie on a circle with radius 1 and centre at (0,0). Naturally, the
couples (x, y) satisfying (i) lie inside the circle with radius 1 and centre (0,0) (you can
also take a test point, say (1/2,1/2): (1/2)2 + (1/2)2 = 1/2 ≤ 1). We also need to
consider (ii). Finally, the domain is the semi-circle as displayed in the gure below.
(b) To nd the stationary points, we need to solve the rst-order conditions:
(
∂g(x,y)
=0
∂x
∂g(x,y)
=0
∂y
(
3x2 − 2x = 0
⇔
−2y = 0
(
x(3x − 2) = 0
⇔
y=0
(
(
x=0
x = 2/3
⇔
or
y=0
y=0
We have found 2 stationary points: (0,0) and (2/3,0). To classify them, we need to
consider the second-order conditions. We calculate the second-order derivatives:
00
00
00
00
00
00
00
gxx
(x, y) = 6x − 2; gyy
(x, y) = −2; gxy
(x, y) = gyx
(x, y) = 0; h(x, y) = gxx
(x, y)gyy
(x, y) − [gxy
(x, y)]2 = −
Evaluate the SOC at the 2 stationary points:
00
00
- at (0,0): gxx
(0, 0) = −2 < 0; gyy
(0, 0) = −2 < 0; and h(0, 0) = 4 > 0. So (0,0) is a
local maximum point.
00
00
(2/3, 0) = −2 < 0; and h(2/3, 0) = 0. So (2/3,0) is a
(2/3, 0) = 0; gyy
- at (2/3,0): gxx
saddle point.
(c) We need to consider the behavior of g on the boundary. We decompose the boundary
into 2 zones:
- zone 1: x = 0, −1 < y < 1. The function g is rewritten as: g(0, y) = 3 − y 2 ; this is
function a quadratic function for −1 ≤ y ≤ 1. Its minimum is reached when y = ±1:
the associated points are (0,-1) and (0,1) and the function equals 2. Its maximum is
reached when y = 0: the associated point is (0,0) and we already know it is a local
6
maximum (function equals 3).
- zone 2: x2 + y 2 = 1 and 1 ≥ x ≥ 0. The function g can be rewritten as: g(x, y) =
3 + x3 − x2 − (1 − x2 ) = 2 + x3 ; this function is strictly increasing for 0 ≤ x ≤ 1. Its
minimum is reached when x = 0: so the point is (0,1) and the function equals 2; and
the maximum is reached when x = 1: so the point is (1,0) and the function equals to 3.
Conclusion: global minimum at (0,1) and (0,-1); global maximum at (0,0) and (1,0).
7