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Mathematics · Advanced university

Stationary point of a positive-definite quadratic

This separable quadratic has positive curvature in each coordinate.

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Method and conditions

This separable quadratic has positive curvature in each coordinate.

∇f = 0; positive-definite Hessian gives a unique global minimum

Worked examples

For f(x,y)=5x² + (-1)x + 2y², find the x-coordinate of the unique global minimizer. Round the final result to 4 decimal places if needed.

Set ∂f/∂x=10x + (-1)=0 and ∂f/∂y=4y=0. The Hessian is positive definite.

0.1

For f(x,y)=3x² + (25)x + 3y², find the x-coordinate of the unique global minimizer. Round the final result to 4 decimal places if needed.

Set ∂f/∂x=6x + (25)=0 and ∂f/∂y=6y=0. The Hessian is positive definite.

-4.1667

Try it

  1. For f(x,y)=11x² + (-19)x + 7y², find the x-coordinate of the unique global minimizer. Round the final result to 4 decimal places if needed.

  2. For f(x,y)=15x² + (-15)x + 15y², find the x-coordinate of the unique global minimizer. Round the final result to 4 decimal places if needed.

Check your work

Answers and marking notes
  1. 0.8636

    Set ∂f/∂x=22x + (-19)=0 and ∂f/∂y=14y=0. The Hessian is positive definite.

  2. 0.5

    Set ∂f/∂x=30x + (-15)=0 and ∂f/∂y=30y=0. The Hessian is positive definite.

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