Multivariate Calculus - Bruce E. Shapiro

114 LECTURE 14. UNCONSTRAINED OPTIMIZATION
Derivation of the Second Derivative Test
Let z = f(x, y) and suppose that P = (a, b) is a critical point where all the partials
are zero, namely,
f x (a, b) = 0
An equivalent way of stating this is that
f y (a, b) = 0
∇f(a, b) = if x (a, b) + jf y (a, b) = 0
We ask the following question: what conditions are sufficient to ensure that P is
the location of either a maximum or a minimum?
We will study this question by expanding the function in a Taylor Series approximation
about a critical point where the partials are zero.
f(x) = f(P ) + f x (P ) (x − a) + f y (P ) (y − b)
+ 1 2 f xx(P ) (x − a) 2 + f xy (P ) (x − a) (y − b) + 1 2 f yy(P ) (y − b) 2
If the first partial derivatives are zero, then
f(x) = f(P ) + 1 2 f xx(P ) (x − a) 2
Next we make a change of variables
Then
+f xy (P ) (x − a) (y − b) + 1 2 f yy(P ) (y − b) 2
u = x − a
v = y − b
g(u, v) = f(x, y) − f(a, b)
∂g
∂u = ∂
∂f ∂x
[f(x, y) − f(P )] =
∂u ∂x ∂u + ∂f ∂y
∂y
and so forth gives us the relationships
∂u = ∂f
∂x
f xx (P ) = g xx (0) = g xx,0
f xy (P ) = g xy (0) = g xy,0
f yy (P ) = g yy (0) = g yy,0
∂f ∂f
(1) + (0) =
∂y ∂x
We have transformed the critical point of f(x, y) at (a, b) to a critical point of g(u, v)
at the origin, where
g(u, v) = 1 2 g uu,0u 2 + g uv,0 uv + 1 2 g vv,0v 2
Revised December 6, 2006. Math 250, Fall 2006