如何在R中寻找函数极值并判断类型？求f(x,y)=x²+y²的简便解法

Great question! For a function as clean as f(x,y) = x² + y², you absolutely don’t need to go through the full workflow of partial derivatives, solving systems with nleqslv, or computing the Hessian. Let’s walk through the easiest shortcuts:

1. Algebraic Non-negativity Argument #

Since squares of real numbers are always non-negative (x² ≥ 0 and y² ≥ 0 for all real x,y), their sum f(x,y) can never be less than 0. The only time f(x,y) hits 0 is when both x=0 and y=0—that’s your extremum right there. And because every other input gives a positive value, this is a global minimum (not just a local one). No calculus required!

2. Geometric Intuition #

Think about what f(x,y) = x² + y² represents: it’s an upward-opening paraboloid (like a bowl sitting perfectly on the origin). The lowest point of the bowl is obviously at (0,0), where the function value is 0. There’s no maximum here because the bowl keeps rising infinitely as you move away from the origin.

3. Quadratic Type Classification (Generalizable) #

This function is a simple quadratic form. For any quadratic function of the form f(x,y) = ax² + bxy + cy² + dx + ey + f, you can:

• First handle the linear terms (dx + ey) by completing the square to find the vertex of the paraboloid/hyperboloid.
• Then check the quadratic part’s definiteness:
  • If the quadratic form is positive definite (like here, where the matrix [[1,0],[0,1]] has all positive eigenvalues), the vertex is a global minimum.
  • If negative definite, it’s a global maximum.
  • If indefinite (mix of positive and negative eigenvalues), there’s no local extremum (just a saddle point).

For f(x,y) = x² + y², there are no linear terms, so the vertex is at (0,0), and the positive definite quadratic part confirms it’s a minimum.

When to Fall Back to Calculus #

These shortcuts work best for symmetric, non-negative, or quadratic functions. For more complex, non-smooth, or arbitrary functions, the partial derivative/Hessian method (and tools like nleqslv) becomes necessary. But for cases like this? Keep it simple!

内容的提问来源于stack exchange，提问作者Alexander Chervov