Questions: Constrained Optimization and Lagrange Multipliers

Question 1

To maximize f(x, y) = xy subject to x + y = 10, you set up ∇f = λ∇g with g(x, y) = x + y - 10. What are the correct gradient components ∇f and ∇g?

• A. ∇f = (y, x) and ∇g = (1, 1)
• B. ∇f = (x, y) and ∇g = (1, 1)
• C. ∇f = (y, x) and ∇g = (x, y)
• D. ∇f = (1, 1) and ∇g = (y, x)

Answer: (A)

The gradient of f(x, y) = xy is (∂f/∂x, ∂f/∂y) = (y, x). The gradient of g(x, y) = x + y - 10 is (∂g/∂x, ∂g/∂y) = (1, 1). Setting ∇f = λ∇g gives the system y = λ and x = λ, meaning x = y. Combined with the constraint x + y = 10, you get x = y = 5 and a maximum of f = 25.

Question 2

If ∇f = λ∇g yields a unique solution point, that point is expected to be the global maximum of f subject to g(x, y) = 0.

• T. True
• F. False

Answer: False

Lagrange's condition ∇f = λ∇g identifies candidates for extrema — points where the constraint curve is tangent to a level curve of f. A unique solution could be a maximum, a minimum, or (in degenerate cases) neither. To determine which, you must evaluate f at all candidate points and compare values, or use second-order conditions. On a compact (closed and bounded) constraint, the maximum and minimum both exist, so the highest and lowest values among candidates are indeed the extrema — but you cannot know which is which without comparing.

Question 3

Give the geometric interpretation of why ∇f and ∇g must be parallel at a constrained optimum.

Model answer: At a constrained optimum, the constraint curve g = 0 must be tangent to a level curve of f. If they crossed instead, you could move along the constraint to reach a higher (or lower) level curve of f, contradicting the assumption that we are at an optimum. Since the gradient of a function is always perpendicular to its level curves, both ∇f and ∇g are perpendicular to the same tangent direction — meaning they must be parallel to each other.

This geometric reasoning is the heart of the Lagrange method. The condition ∇f = λ∇g is not an arbitrary algebraic trick; it captures exactly the geometric situation where you cannot improve f by moving along the constraint. The scalar λ tells you how fast the optimal value of f changes if you relax the constraint — making it a powerful economic and physical interpretive tool (e.g., in resource allocation, λ is the 'shadow price' of the constraint).