Constrained Optimisation
Course Manual 10.1 and 10.2
Jacques 5.5 and 5.6

The behaviour of economic actors is often constrained by the economic resources they have at their disposal

Example:

–Individuals maximising utility will be subject to a budget constraint

–Firms maximising output will be subject to a cost constraint

The function we want to maximise is called the objective function

The restriction is called the constraint

Max U = f(X1 X2) = X21 X2

[ X1* X2* ]

Subject to

g(X1 X2)= P1X1 + P2X2 - M = 0

Three Ways to do this:

  • By Substitution
  • Economic Theory
  • Lagrange Multiplier

Method 1: By Substitution

Max the Objective function:

Max U = f (X1 X2) = X21 X2

[ X1* X2* ]

Subject to the constraint:

P1X1 + P2X2= M

Step 1:Use the constraint to express X2 in terms of X1 (or vice-versa)

Step 2:

Substitute into objective function

Step 3:

F.O. Condition

( P1X1 = 2/3M, expenditure on good 1 is 2/3 of income)

S. O. Condition

For a Max,

X1 needs to be large enough to sign N.D.

How Large? First find the X1 that sets,

Answer:

The optimal f11< 0

Step 4: Substitute into constraint to find corresponding value of X*2 that maximises objective function

Since P1X1 + P2X2= M

(note, rearranging, P2X2 = 1/3 M . expenditure on good 2 is 1/3 of M)

Method 2: By Economic Theory

Max U = f (X1 X2) =

[ X1* X2* ]

S.t. g (X1 X2)= P1X1 + P2X2 - M =0

  • Total differential of the objective function f (X1 X2):

dU = f1 dX1+ f2 dX2 = 0

(f1f2, marginal utility of X1X2, respectively)

eq.1

  • Total differential of the constraint:

g(X1X2) = P1 X1+ P2 X2 – M= 0

P1 dX1+ P2 dX2 = 0

eq.2

Condition 1 – from eq1 and eq2

Condition 2

From condition 1, substitute P1X1 (or P2X2) into the constraint, and solve

Hence,

Method 3: By The Lagrange Multiplier

Max U = f (X1 X2) =

[ X1* X2* ]

s.t.

g (X1 X2)= P1X1 + P2X2- M = 0

Step 1: Define the Lagrangean Function L (objective function +  constraint)

Max L = f(X1 X2 ) + g(X1 X2)

[X1* X2* *]

Max L = X12X2 + (M – P1X1– P2X2 )

[X1* X2* *]

OR L = X21X2 – (P1X1 + P2X2 –M)

Step 2: First Order Conditions

Set dL= L1.dX1 + L2.dX2 = 0

1. L1 = 2X1X2 –  P1 = 0 eq1

2. L2 = X12 –  P2 = 0 eq2

Set g(X1X2) = 0

3. L = M – P1X1 – P2X2 = 0 eq3

Step 3: Solve the system of equations

Solving equations 1 & 2:

 2X1P2X2 = P1X12

 2P2X2 = P1X1

And substituting into eq 3

P1X1 + P2X2 – M= 0

2P2X2+ P2X2 – M= 0

from eq 3:

Substituting in for X2:

(again, note that rearranging reveals that P1X! = 2/3 M and P2X2 = 1/3 M .2/3 of income spent on good 1, and 1/3 on good2)

Step 4: Second Order Condition

d2L = L11.dX21+L12.dX1dX2+L21.dX2dX1+L22dX22

s.t. g1.dX1 + g2.dX2= 0

or dX2= -(g1/ g2).dX1

N. D. for a Max

d2L=[L11.g22 - 2L12.g1.g2+L22g21]dX21/g22

d2L= ΦdX21/g22 , if Φ < 0, N. D.

= 0+ P1(P2.2X1) -P2(-P1.2X1+ P2.2X2)

= 4P1P2X*1 - 2P2P2.X*2

= 4P2(2/3. M) - 2P2(1/3.M)

= (8P2M - 2P2M)/3

= 2P2M > 0

d2L < 0, N. D. (Max)

A consumers preferences can be represented by the Utility Function, U(x,y)=x.y. How much will the utility maximising consumer demand of goods x and y if they have an income of €100, the price of good x is €5 and the price of good y is €1?

Lagrangean Method:

L = x.y + [100 – 5x – y]

Eq. 1. LX = y – 5= 0

Eq. 2. Ly = x –  = 0

Eq. 3. L =100 – 5x – y = 0

Solving

Eq 1&2:  = y/5 = x

So y* = 5x

Substitute into eq.3:

100 = 5x + y = 10x

So x* = 10

Then y*= 5x=50

And U* = xy = 500

(note that P1X1 = ½ M and P2X2 = ½ M .

½ of income spent on good 1, and ½ on good2)

Second Order Condition:

Topic 10: Q1

Maximise (i) U = x½y½ (ii) U = x.y (iii) U = logx + logy. Let price x = p and price of y = q, and m = income

(i) L = x½y½ + [m – px – qy]

Eq. 1. Lx = ½ x-½y½ - p= 0

Eq. 2. Ly = ½ x½y-½ - q = 0

Eq. 3. L = m – px – qy = 0

Eq 1&2:  =

 px½y-½ = qx-½y½

 p x½/ x-½ = q y½/ y-½

 px = qy….expenditure on good x = expenditure on good y

Substitute into eq.3: m = px + qy = 2px

px* = ½ m qy* = ½ m

Cobb-Douglas: if U = xayb and a+b=1 (a= ½ & b= ½ above)Then income expenditure share on x is a and on y is b.

(ii) L = xy + [m – px – qy]

Same Solution: x* = m/2p (and px = ½ m) and y*= m/2q (and qy = ½ m)

Cobb-Douglas: if U = xayb and a+b1 (a=1 & b=1 above)

Then income expenditure share on x is a/a+b and on y is b/a+b

(iii) L = log x + log y + [m – px – qy]

Same Solution: x* = m/2p (and px = ½ m) and y*= m/2q (and qy = ½ m)

Note: log x + log y = x.y

3 utility functions - same ranking of preferences – represent monotonic transformations of a utility function.