Unit 4: Differential Calculus - 2 (Multi-variable case)

1. Partial and Total Derivative

Most economic models involve more than one variable. For example, Utility depends on goods X and Y: U = f(X, Y). A Production Function depends on Labor (L) and Capital (K): Q = f(L, K).

Partial Derivative

Definition: A partial derivative measures the rate of change of a multi-variable function with respect to one variable, while holding all other variables constant.

• The partial derivative of f(X, Y) with respect to X is ∂f/∂X or f_X.
• The partial derivative of f(X, Y) with respect to Y is ∂f/∂Y or f_Y.

Example: Let U(X, Y) = X²Y³
To find ∂U/∂X, treat Y³ as a constant: ∂U/∂X = (Y³) * (2X) = 2XY³
To find ∂U/∂Y, treat X² as a constant: ∂U/∂Y = (X²) * (3Y²) = 3X²Y²

Total Derivative and Total Differential

The total differential (df) measures the total change in the function f(X, Y) resulting from small changes in both X (by dX) and Y (by dY) simultaneously.

df = (∂f/∂X)dX + (∂f/∂Y)dY

Total Change = (Change from X) + (Change from Y)

2. Unconstrained Optimisation problem

This is the process of finding the maximum or minimum of a function of two or more variables, like f(X, Y), without any constraints.

Example: A firm wants to maximize profit π(L, K) = P*Q(L, K) - wL - rK by choosing the optimal amount of Labor (L) and Capital (K).

Steps for Unconstrained Optimization:

1. First-Order Condition (FOC):

   Find the critical point (X₀, Y₀) where the function is "flat" in all directions. This means all partial derivatives must be zero simultaneously.

   ∂f/∂X = 0 ∂f/∂Y = 0

   You must solve this system of two equations for X and Y.

2. Second-Order Condition (SOC):

   This test determines if the critical point is a max, min, or saddle point. It uses the Hessian determinant (see section 5).

3. Constraint optimisation

This is the most common type of problem in economics. It involves maximizing or minimizing an objective function subject to a constraint (which limits your choices).

Common Economic Examples:

• Problem 1: Consumer's Choice
  • Objective: Maximize Utility, U(X, Y)
  • Constraint: Budget, P_X*X + P_Y*Y = M
• Problem 2: Producer's Choice (Cost Minimization)
  • Objective: Minimize Cost, C(L, K) = wL + rK
  • Constraint: Production Quota, Q(L, K) = Q₀

The primary method to solve these problems is the Lagrangian multiplier method.

4. Lagrangian multiplier

The Lagrangian method combines the objective function and the constraint into a new, unconstrained function called the Lagrangian (L).

For a general problem:
Maximize f(X, Y)
Subject to g(X, Y) = c

Steps to use the Lagrangian Method:

1. Set up the Lagrangian Function (L):

   L includes the objective function, the constraint, and a new variable λ (lambda), the Lagrangian multiplier.

   L(X, Y, λ) = f(X, Y) - λ [g(X, Y) - c]

   (Note: Some use +λ, which just flips the sign of λ's interpretation. Both are valid if used consistently.)

2. Find the First-Order Conditions (FOCs):

   Take the partial derivative of L with respect to all variables (X, Y, and λ) and set them all to zero.

   (1) ∂L/∂X = (∂f/∂X) - λ(∂g/∂X) = 0 (2) ∂L/∂Y = (∂f/∂Y) - λ(∂g/∂Y) = 0 (3) ∂L/∂λ = -[g(X, Y) - c] = 0 => g(X, Y) = c

   Notice that FOC (3) simply restores our original constraint!

3. Solve the System:

   You now have a system of 3 equations and 3 unknowns (X, Y, λ). Solve them to find the optimal values X*, Y*, and λ*.

5. Hessian determinants

The Hessian is a matrix of all the second-order partial derivatives. It is the multi-variable version of the "second derivative" and is used in the SOC for unconstrained optimization.

For a function f(X, Y), we have four second-order partials:

• f_XX = ∂²f/∂X² (Differentiate wrt X, then wrt X again)
• f_YY = ∂²f/∂Y² (Differentiate wrt Y, then wrt Y again)
• f_XY = ∂²f/∂Y∂X (Differentiate wrt X, then wrt Y)
• f_YX = ∂²f/∂X∂Y (Differentiate wrt Y, then wrt X)

Young's Theorem: As long as the second partials are continuous, f_XY = f_YX.

The Hessian Matrix (H):

H = [ f_XX f_XY ]
    [ f_YX f_YY ]

The Hessian determinant |H| is: |H| = (f_XX)(f_YY) - (f_XY)²

Second-Order Condition (SOC) for Unconstrained Optimization:

At a critical point (where f_X = 0 and f_Y = 0):

• Local Maximum: f_XX < 0 AND |H| > 0
• Local Minimum: f_XX > 0 AND |H| > 0
• Saddle Point: |H| < 0
• Inconclusive: |H| = 0

6. Bordered Hessian determinant

The Bordered Hessian is the SOC for constrained optimization. It is the Hessian of the Lagrangian function, "bordered" by the derivatives of the constraint.

For L(X, Y, λ) with constraint g(X, Y) = c:
g_X = ∂g/∂X and g_Y = ∂g/∂Y
L_XX = ∂²L/∂X², L_XY = ∂²L/∂Y∂X, etc.

The Bordered Hessian Matrix (|H_B|):

|H_B| = [ 0 g_X g_Y ]
          [ g_X L_XX L_XY ]
          [ g_Y L_YX L_YY ]

Second-Order Condition (SOC) for Constrained Optimization:

At the optimal point (X*, Y*, λ*):

• Local Maximum: |H_B| > 0
• Local Minimum: |H_B| < 0

7. Economic application

This entire unit is about the core mathematical tools of microeconomics.

Example: Utility Maximization (Full Problem)

• Problem: Maximize U = XY subject to 2X + 4Y = 120.
• 1. Lagrangian:
  L = XY - λ(2X + 4Y - 120)
• 2. FOCs:
  (1) ∂L/∂X = Y - 2λ = 0 => Y = 2λ
  (2) ∂L/∂Y = X - 4λ = 0 => X = 4λ
  (3) ∂L/∂λ = -(2X + 4Y - 120) = 0 => 2X + 4Y = 120
• 3. Solve:
  Substitute (1) and (2) into (3):
  2(4λ) + 4(2λ) = 120
  8λ + 8λ = 120
  16λ = 120 => λ* = 7.5
  
  Now find X* and Y*:
  X* = 4λ = 4(7.5) = 30
  Y* = 2λ = 2(7.5) = 15
  
  Optimal Bundle: (X=30, Y=15)
  Interpretation of λ: If income increases by1 (to 121), utility will increase by approx. 7.5.
• 4. Check SOC (Bordered Hessian):
  g_X = 2, g_Y = 4
  L_XX = 0, L_YY = 0, L_XY = 1
  |H_B| = [ 0 2 4 ]
            [ 2 0 1 ]
            [ 4 1 0 ]
  
  |H_B| = 0(0-1) - 2(0-4) + 4(2-0) = 0 - 2(-4) + 4(2) = 8 + 8 = 16
  Since |H_B| = 16 > 0, we have confirmed this is a maximum.