ECON 331 - Mathematical Economics - ANSWERS FINAL EXAM

1. (a)- The consumer's utility maximization problem is written as:

maxx,y

[xy + y2 + 2x + 2y] s.t. 6x + 10y = m, x ≥ 0, y ≥ 0

- The associated Lagrangian function is de�ned as:

L(x, y, λ) = xy + y2 + 2x + 2y − λ [6x + 10y −m]

where λ is the Lagrange multiplier.- The Kuhn-Tucker conditions are written as:

(i) x∂L(x, y, λ)

∂x= 0, x ≥ 0,

∂L(x, y, λ)

∂x= y + 2− 6λ ≤ 0

(ii) y∂L(x, y, λ)

∂y= 0, y ≥ 0,

∂L(x, y, λ)

∂y= x + 2y + 2− 10λ ≤ 0

(iii)∂L(x, y, λ)

∂λ= − [6x + 10y −m] = 0

- From (iii): x = m/6− 5/3y (*).- Suppose y = 0. From (*) x = m/6 > 0 ⇒ ∂L

∂x= 0 ⇒ λ = 1/3. Plug λ and x in (ii):

you �nd ∂L∂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

∂y= 0 ⇒ x + 2y + 2− 10λ = 0 (**).

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

∂x= −m/50 + 4/5 ≤ 0. However, remember that

8 < m < 40 ⇒ 0 < −m/50 + 4/5 which is a contradiction.So suppose instead x > 0 ⇒ ∂L

∂x= 0 ⇔ y = 6λ−2. And from (*) x = m/6−10λ+10/3.

Plug x and y in (**) and solve for λ. You �nd:

x∗ = (40−m)/24 > 0 b/c m < 40

y∗ = (m− 8)/8 > 0 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 withone choice variable, x and one non-negativity constraint.

(b) We have seen a result in class that states:

∂U(x∗, y∗)∂m

= λ

1

We check by calculation that it works here:

U(x∗, y∗) =40−m

24× m− 8

8+

(m− 8

8

)2

+40−m

12+

m− 8

4=

1

96m2 +

1

6m +

2

3

⇒ ∂U(x∗, y∗)∂m

=1

48m +

1

6(= λ)

⇒ ∂U(x∗, y∗)∂m |m=20

=7

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(c) - case where 0 ≤ m ≤ 8: previous solution does not work because it violates thepositivity of x∗. So we need to study the behavior of U on the boundary.. x = 0 and y > 0: U(0, y) = y2 + 2y with y > 0 and 10y = m. So y = m/10 andU1 = 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∗ (foundin b)) is violated:. same 2 cases and associated solutions! This time, U1 > U2 (when m ≥ 40), hence thesolution is x∗ = 0 and y∗ = m/10.

2. (a)

In + sM = (In + M)−1 ⇔ (In + sM)(In + M) = In

⇔ In + M + sM + sM2 = In

⇔ (1 + s)M + s(3M) = 0 b/c by assumption M2 = 3M

⇔ (1 + 4s)M = 0

⇔ 1 + 4s = 0 or M = 0

⇔ s = −1

4or M = 0

To conclude, for any matrix M s.t. M2 = 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 b/c B is invertible by ass.⇔ X = (A− I)−1CB−1 b/c A− In is invertible by ass.

(ii)

A−In =

(1 1

1 −1

); (A−In)−1 =

(1/2 1/2

1/2 −1/2

)apply the inverse formula for (2,2)-matrices

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and B−1 =

1 0 0

0 2 0

0 0 4

apply the inverse formula for diagonal matrices

X =

(1 1

1 −1

)×

(1 1 2

−1 3 1

)×

1 0 0

0 2 0

0 0 4

=

(0 2 3/2

1 −1 1/2

)×

1 0 0

0 2 0

0 0 4

=

(0 4 6

1 −2 2

)

3. - Taylor's approximation: we de�ne a function and Taylor provides a polynomialapproximation of this function around a point of interest. Here I de�ne f(x) = x1/3

and I want to approximate it around x0 = 8. The second-order approximation is:

f(x) ∼ f(8) + f ′(8)(x− 8) +1

2f ′′(8)(x− 8)2

f(8.1) ∼ 2 +1

38−2/3 × (8.1− 8)− 1

2× 2× 8−5/3

3× 3× (8.1− 8)2

= 2 +1

12× 0.1− 1

144× 2× 0.12

= 2 + 0.008333− 0.000035

= 2.0083

- Newton's algorithm: we de�ne an equation that has solution at x = 8.11/3 andNewton's provides an approximate solution. Here I de�ne g(x) = x3 − 8.1 and theequation g(x) = 0 has solution at x = 8.11/3. The algorithm is given by:

xi+1 = xi − g(xi)

g′(xi)(whenever g′ 6= 0)

x1 = 2− 8− 8.1

3× 22=

241

120∼ 2.00833

x2 =241

120− (241/120)3 − 8.1

3× (241/120)2∼ 2.0083

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4. (a) - We can build up the following table of revenues for successive years:

Y ear Revenue

0 A

1 A− pA = (1− p)A

2 (1− p)A− p[(1− p)A] = (1− p)A[1− p] = (1− p)2A... ...n (1− p)nA

So the revenue in year n can be expressed as Rn = (1− p)nA. (formula can be provedby induction - not requested here.)- The company operates as long as the yearly revenue exceeds the yearly cost, ie as longas (1 − p)nA > K. So we need to �nd the largest integer n∗ (representing the largestnumber of years) that satis�es this equation.

(1− p)nA > K ⇔ ln [(1− p)nA] > ln(K) (K > 0!!)

⇔ n ln(1− p) + ln(A) > ln(K)

⇔ n ln(1− p) > ln(K)− ln(A)

⇔ n ln(1− p) > ln

(K

A

)

⇔ n <ln

(KA

)

ln(1− p)(0 < (1− p) < 1, so ln(1− p) < 0, so inequality changes!!)

So n∗ is the largest integer satisfying the above equation.(b) The total pro�t from year 0 to n∗ is simply the di�erence between the total revenueand the total cost from year 0 to n∗. The total revenue is the sum of the yearly pro�tsfrom year 0 to n∗, ie:

R =n∗∑i=0

Ri =n∗∑i=0

[(1− p)iA] =

[n∗∑i=0

(1− p)i

]A

=1− (1− p)n∗+1

1− (1− p)A ( use the hint with x = 1− p)

=1− (1− p)n∗+1

pA

The cost is the same each year, so the total cost is simply: C = (n∗ + 1)×K

Finally, the total pro�t is:

π =1− (1− p)n∗+1

pA− (n∗ + 1)×K

(c) (i) Plug the given values in the equation found in (a) to get:ln

(KA

)

ln(1− p)=

ln(

50000007000000

)

ln(0.98)= 16.65

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n∗ is the largest integer that is smaller than 16.65, so n∗ = 16.(ii) The associated pro�t is:

π =1− (0.98)17

0.02× 7000000− 17× 5000000 = 16.74× 106

(with n∗ = 14, we get: π = 6.5× 106).

5. (a) The coe�cient matrix of the system is:

A =

1 3 4

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):

det(A) =

∣∣∣∣∣2 1

−3 −9

∣∣∣∣∣− 2

∣∣∣∣∣3 4

−3 −9

∣∣∣∣∣ + 3

∣∣∣∣∣3 4

2 1

∣∣∣∣∣= −18 + 3− 2(−27 + 12) + 3(3− 8)

= 0

(b)(i) From (a), the coe�cients 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 in�-nite number of solutions, but we cannot tell yet. It cannot be solved by matrix-inversionand 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)

⇔

x + 3y + 4z = b1

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 coe�cients b1, b2, and b3 satisfy the lastequation: 3b1 − 3b2 + b3 = 0.

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6. (a) The domain is de�ned by couples (x, y) that satisfy 2 equations: (i) x2 +y2 ≤1; (ii) x ≥ 0. To sketch the domain, we use the hint! We know that couples (x, y)

satisfying x2 + y2 = 1 lie on a circle with radius 1 and centre at (0,0). Naturally, thecouples (x, y) satisfying (i) lie inside the circle with radius 1 and centre (0,0) (you canalso take a test point, say (1/2,1/2): (1/2)2 + (1/2)2 = 1/2 ≤ 1). We also need toconsider (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)∂x

= 0∂g(x,y)

∂y= 0

⇔{

3x2 − 2x = 0

−2y = 0

⇔{

x(3x− 2) = 0

y = 0

⇔{

x = 0

y = 0or

{x = 2/3

y = 0

We have found 2 stationary points: (0,0) and (2/3,0). To classify them, we need toconsider the second-order conditions. We calculate the second-order derivatives:

g′′xx(x, y) = 6x− 2; g′′yy(x, y) = −2; g′′xy(x, y) = g′′yx(x, y) = 0; h(x, y) = g′′xx(x, y)g′′yy(x, y)− [g′′xy(x, y)]2 = −12x + 4

Evaluate the SOC at the 2 stationary points:- at (0,0): g′′xx(0, 0) = −2 < 0; g′′yy(0, 0) = −2 < 0; and h(0, 0) = 4 > 0. So (0,0) is alocal maximum point.- at (2/3,0): g′′xx(2/3, 0) = 0; g′′yy(2/3, 0) = −2 < 0; and h(2/3, 0) = 0. So (2/3,0) is asaddle point.(c) We need to consider the behavior of g on the boundary. We decompose the boundaryinto 2 zones:- zone 1: x = 0, −1 < y < 1. The function g is rewritten as: g(0, y) = 3 − y2; this isfunction 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 isreached when y = 0: the associated point is (0,0) and we already know it is a local

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maximum (function equals 3).- zone 2: x2 + y2 = 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. Itsminimum is reached when x = 0: so the point is (0,1) and the function equals 2; andthe 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).

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