Part a) Lagrange Problem and Envelope Theorem
Given:
- Maximize:
f(u, v) = u + 3v - Subject to:
g(u, v) = u² + av² = 10 - We are told to apply the Envelope Theorem and estimate f* when a = 1.01
Step 1: Lagrangian Function
L(u, v, λ) = u + 3v + λ(10 − u² − av²)
Step 2: First-Order Conditions
∂L/∂u = 1 − 2λu = 0 → λ = 1/(2u)∂L/∂v = 3 − 2λav = 0 → λ = 3/(2av)- Equating both expressions for λ:
1/(2u) = 3/(2av) → v = 3au
Step 3: Plug into Constraint
u² + a(3au)² = 10
u² + 9a²u² = 10 → u²(1 + 9a²) = 10
u² = 10 / (1 + 9a²) → u = √[10 / (1 + 9a²)]
Step 4: Calculate f*(a)
v = 3au → f(u,v) = u + 3v = u + 3(3au) = u(1 + 9a)
So, f*(a) = √[10 / (1 + 9a²)] * (1 + 9a)
Step 5: Estimate f*(1.01)
a = 1.01 → a² = 1.0201Denominator = 1 + 9(1.0201) = 10.1809Numerator = 1 + 9(1.01) = 10.09f*(1.01) ≈ √(10 / 10.1809) * 10.09√(0.9822) ≈ 0.99106- Estimated f* ≈ 0.99106 × 10.09 ≈ 10.00
Envelope Theorem Check:
Envelope Theorem tells us:
df*/da = −λ * ∂g/∂a = −λ * v²
Since v = 3au and u = √[10 / (1 + 9a²)], this can be verified by substitution if exact change is required.
Part b) Maximize f(u;a) = −u² + 2au + 4a
Given:
- f(u; a) = −u² + 2au + 4a
- We need to maximize over u, where a > 0
Step 1: First Order Condition
∂f/∂u = −2u + 2a = 0 → u = a
Step 2: Plug back into f(u; a)
f(a; a) = −a² + 2a² + 4a = a² + 4a
Thus, optimal value of function:
f* = a² + 4a
Implications:
- The function increases with increasing a
- As a → ∞, f* → ∞
- So maximum increases with a for a > 0
Conclusion
We solved two Lagrange problems — one using Envelope Theorem to estimate the optimal value with a slight change in parameter a, and the second by directly optimizing a function with respect to u. These techniques are common in economic optimization and sensitivity analysis. Envelope Theorem helps simplify comparative statics when a constraint changes slightly, while direct optimization gives closed-form solutions under known conditions.