f(x,y) = x^2 + kxy + y^2 where k is a constant.

Question

For what values of k does the second derivative test guarantee a saddle point.

For what values of k does the second derivative test guarantee a local minimum.

Answer 1

If we can show that there is a point where the gradient is (0, 0) and the matrix of second derivatives is positive definite, then there is a local minimum there. If the matrix of second derivatives is positive definite everywhere, then this is also a global minimum.

With this problem, the gradient is (2x + ky, 2y + kx), which equals (0, 0) at (0, 0)

The matrix of second derivatives is

2 k

k 2

Solving for eigenvalues, (2 - λ)² - k² = 0

λ = 2 - k, 2 + k

If |k| < 2, both 2 - k and 2 + k are positive, so the matrix is positive definite and we have a local (and global) minimum at (0, 0)

If |k| > 2, either 2 + k or 2 - k is negative, so we have a saddle point.

If k = 2, the second derivative test does not guarantee a local minimum or saddle point.

Of course, it turns out that, if k = 2, the function may be written f(x,y) = (x - y)², which has a global minimum at x = y.