Lecture 7
CONSTRAINED OPTIMIZATION

1. Objective

• Objective function: z = f (x,y)
• Constraint: g(x,y) = c

  [Note: The constraint may be written as: c - g(x,y) = 0.]

2. The Lagrange-multiplier method

• Form the Lagrangian function:
  
  L(x,y,λ) = f(x,y) + λ[c - g(x,y)]
  
  
• λ is the Lagrange multiplier
• Treat the Lagrangian function as the new objective function, with the choice variables as x, y and λ.

2.1 First-order conditions

Use the FOC to obtain the extrema of the function.

• The first-order necessary conditions are:
  
  
  L₁ = f₁(x,y) - λg₁(x,y) = 0       (1)

  L₂ = f₂(x,y) - λg₂(x,y) = 0       (2)

  L₃ = c - g(x,y) = 0                     (3)

  [Note: Equation (3) gives us back the constraint!]

• Solve equations (1)-(3) for x, y and λ.
• These values of x and y will yield an extremum for f(x,y).

2.2 Example

Find the extremum of the function z = xy, subject to the constraint x + y = 8.

1. Obtain the Lagrangian function, L(x,y,λ).
2. Write down the first-order conditions (3 of them).
3. Solve the three equations for the three variables.
4. Obtain the stationary value of z.

   [Note: At this point we do not know if the extremum is a maximum or a minimum. We will develop the SOC later.]

2.3 Second-order conditions

The second-order conditions entail the use of:

• The second-order partial derivatives of L(x,y,λ), and
• The first-order derivatives of g(x,y).
  

3. The Bordered Hessian

• Evaluate the partial derivatives--L₁₁, L₁₂, L_21, L₂₂--at the extremum.
• Form a determinant with the partial derivatives, and border it on two sides by g₁ and g₂.
• The bordered Hessian (H_bar) is:

  			0	g1	g2
  	H_bar = 	g1	L11	L12
  			g2	L21	L22

• Sufficient condition for a maximum: det(H_bar) > 0
• Sufficient condition for a minimum: det(H_bar) < 0

4. The n-variable case

• Objective function: z = f(x₁, x₂,..., x_n)
• Constraint: g(x₁, x₂,..., x_n) = c

1. Form the Lagrangian function:

   L(x₁, x₂,..., x_n, λ) = f(x₁, x₂,..., x_n) + λ[c - g(x₁, x₂,..., x_n)]

2. Obtain the FOC:

   	L1(x1, x2,..., xn, λ) = 0		(1)
   	L2(x1, x2,..., xn, λ) = 0		(2)
   	...
   	Ln(x1, x2,..., xn, λ) = 0		(n)
   	Lλ(x1, x2,..., xn, λ) = 0		(n+1)
   

3. Solve equations (1), (2),...,(n+1) to obtain x₁, x₂,..., x_n and λ.
4. Obtain the bordered Hessian:
   
   H_bar =

   		0	g1	g2	...	gn
   		g1	L11	L12	...	L1n
   		g2	L21	L22	...	L2n
   		...
   		gn	Ln1	Ln2	...	Lnn
   

5. Obtain the bordered principal minors:

   			0	g1	g2
   	H2_bar = 	g1	L11	L12
   			g2	L21	L22
   
   
   	H3_bar, H4_bar,..., Hn_bar
      

6. Check the second-order conditions (SOC)
   
   
   For a maximum: H₂_bar > 0, H₃_bar < 0, ...

   For a minimum: H₂_bar < 0, H₃_bar < 0, ...

5. Interpretation of the Lagrange multiplier

• The multiplier measures the change in the optimal value of the objective function due to a change in the parameter c in the constraint, i.e.
  
  
  λ = df^*/dc

6. Application: Utility maximization

	max 		u(x,y)
	x,y

	s.t. 		pxx + pyy = I

• The optimal bundle is obtained at the tangency between the indifference curve and the budget constraint.

1. Form the Lagrangian function.
2. Obtain the FOC.
3. Obtain the bordered Hessian.
4. What restrictions on u(x,y) will ensure that the FOC will yield a maximum?
5. What is the interpretation of the multiplier?

6.1 Example

Consider the utility function u(x,y) = (x+2)(y+1). The prices of the two goods and the consumer's income are: p_x = 4, p_y = 6, I = 130.

1. Solve for the optimal levels of x and y.
2. Are the second-order conditions for a maximum satisfied?
3. What will happen to the optimal levels of x and y in response to an increase in the price of Good X? How about utility? (Assume p_x increases by 10%.)
4. What will happen to the optimal levels of x and y in response to an increase in income? What will ahppen to utility? (Assume I increases to 150.)