Mathematics for Economics

Question one:

S=100U+310E-½U2–2E2–UE

First order condition, differentiation:

dS/dU = 100 – 2U – E = 0

dS/dE = 310 – 4E – U = 0

From the first equation U = 50 – ½ E

From second equation U = 310 – 4E

We solve for E

50-½E=310–4E

E = 74.28

Mathematics for Economics

We now solve for U

U = 50 – ½ (74.28)

U = 12.86

Second order condition:

S11 = dS/dU2 = – 2U – 1

S22 = dS/dE2 = – 4E – 1

S12 = -1

condition S11 . S22 > S12

S11.S22=6/UE

Therefore S11.S22 > -1

Mathematics for Economics

Question 2:

Utility maximization:

Max U (A,B) S.T PaA + PbB = I

L (A,B,λ) = A1/3 B 2/3 – λ (PaA + PbB – I )

First order condition:

dL/ dA = 1/3 A-2/3 B2/3 – λ Pa = 0

dL/ dB = 2/3 A1/3 B-1/3 – λ Pa = 0

dL/ d λ = – (PaA + PbB – I ) = 0

Make λ the subject of the formula:

λ = [ 1/3 A-2/3 B2/3 ] /Pa

λ = [2/3 A1/3 B-1/3] / Pb

[2/3 A1/3 B-1/3] / Pb = [ 1/3 A-2/3 B2/3 ] /Pa

Mathematics for Economics

Pa[2 A1/3 B-1/3] = Pb [ A-2/3B2/3 ]

And that Pa = \$1 and Pb = \$1

2 A1/3 B-1/3 = A-2/3 B2/3

Simplifications of the formula give us

B=2A

And our budget is

PaA + PbB = I

Where I is 300

A+B=300

Substitute B

3A = 300

Mathematics for Economics

A=100

B=200

Therefore to maximize utility we purchase 100 apples and 200 bananas

Question 3:

The above chart shows the efficient frontier

Efficient frontier

Value of the langrage multiplier when portfolio return is 20%

The value of the langrage multiplier is 0.060475

Mathematics for Economics

## REFERENCE:

Michael Hoy (2005) Mathematics for Economics, McGraw Publishers, New  York