Question

Explain, why the maximum-minimum test for quadratic functions is a special case of the second derivative test.

Answer 1

Consider the quadratic function

g(x, y) = Ax² + 2Bxy + Cy² .

The only critical point is (0, 0) and the second derivatives of g(x, y) are

g_xx = 2A

 g_xy = 2B

g_yy = 2C .

The second derivative test states that if at (0, 0) we have g_xx < 0 and g_xx g_yy − g²_xy > 0, then (0, 0) is a local minimum. For our function g(x, y) this means

2A < 0 and 4AC − 4B² > 0 .

When we divive the first inequality by 2 and the second one by 4, we recover the maximum-minimum test for quadratic functions. The same argument can be used for the other types of critical points.

Note: There is one slight difference between the two tests though. The second derivative test tells us when a critical point is a local maximum or minimum. For a quadratic function it is true that a local maximum or minimum is in fact a global one.