Hence, if A> 0 , then A(AC −B^2 ) is negative and alternatively if A< 0 , then
A(AC −B^2 )is positive. In either case, the second derivative has a nonconstant
sign value and in this situation the critical point (x 0 ,y 0 )is said to correspond to
a saddle point. Such a critical point is illustrated in figure 7-15.
3. If AC −B^2 =zxx zyy −(zxy)^2 > 0 ,the second derivative is of constant sign, which is
the sign of A.
(a) If A> 0 , d
(^2) z
ds^2 >^0 , the curve z=z(s)is concave upward for all α, and hence
the critical point corresponds to a relative minimum.
(b) If A< 0 , d
(^2) z
ds^2 <^0 , the curve z =z(s) is concave downward for all α, and
therefore the critical point corresponds to a relative maximum.
Example7-24. Find the maximum and minimum values of
z=z(x,y ) = x^2 +y^2 − 2 x+ 4 y
Solution: The first and second partial derivatives of z are
∂z
∂x
= 2x− 2 , ∂z
∂y
= 2y+ 4 ,A =∂
(^2) z
∂x^2
= 2,C =∂
(^2) z
∂y^2
= 2,B = ∂
(^2) z
∂x∂y
= 0
Setting the first partial derivatives equal to zero and solving for x and y gives the
critical points. For this example there is only one critical point which occurs at
(x 0 ,y 0 ) = (1,−2).From the second derivatives of z one finds A= 2 ,B = 0 ,C = 2 and
AC −B^2 = 4 > 0 ,and consequently the critical point (1 ,−2) corresponds to a relative
minimum of the function.
The use of level curves to analyze complicated surfaces is sometimes helpful. For
example, the level curves of the above function can be expressed in the form
z=z(x,y ) = (x−1)^2 + (y+ 2)^2 −5 = k=constant.
By assigning values to the constant kone can determine the general character of the
surface. It is left as an exercise to show these level curves are circles which are cross
section of the surface known as a paraboloid.
Lagrange Multipliers
Consider the problem of finding stationary values associated with a function
f=f(x,y )subject to a constraint condition that g=g(x,y ) = 0 .Recall that a necessary
