Extremes of x² + y² on the hyperbola xy = 1
Optimising x^2+y^2 subject to x*y = 1 gives 2 at (-1, -1), with multiplier λ = 2. Full worked solution and an editable calculator.
Optimising f = x^2+y^2 subject to x*y = 1 gives f = 2 at (-1, -1), with λ = 2. Finds the two points on the hyperbola closest to the origin.
The quantity to maximise or minimise
Left-hand side of g = c
Right-hand side
The system to solve
∂/∂x: ∂f/∂x = λ · ∂g/∂x ∂/∂y: ∂f/∂y = λ · ∂g/∂y constraint: x*y = 1
Finds the two points on the hyperbola closest to the origin.
Critical points
| x | y | λ | f | Type |
|---|---|---|---|---|
| -1 | -1 | 2 | 2 | maximum |
| 1 | 1 | 2 | 2 | maximum |
More Lagrange multiplier problems
Frequently asked questions
What is the answer to "Extremes of x² + y² on the hyperbola xy = 1"? ▾
The optimum is f = 2 at (-1, -1), with Lagrange multiplier λ = 2. Finds the two points on the hyperbola closest to the origin.
What system of equations does this produce? ▾
Setting ∇f = λ∇g gives one equation per variable (x, y), plus the constraint x*y = 1. That is 3 equations in 3 unknowns, counting λ.
What does λ mean here? ▾
λ = 2 means that raising the constraint value from 1 by one unit would change the optimal value of f by roughly 2.
About this Calculator
Optimising x^2+y^2 subject to x*y = 1 gives 2 at (-1, -1), with multiplier λ = 2. Full worked solution and an editable calculator.