Michael Corral: Vector Calculus

86 CHAPTER 2. FUNCTIONS OF SEVERAL VARIABLES
then the critical points (x,y) are the common solutions of the equations
y−3x 2 = 0
x−2y=0
The first equation yields y=3x 2 , substituting that into the second equation yields
x−6x 2 = 0, which has the solutions x=0 and x= 1 6
. So x=0⇒y=3(0)=0 and
x= 1 6 ⇒ y=3( 1 2
6)
= 1 12 .
So the critical points are (x,y)=(0,0) and (x,y)= ( 1
6 , 12) 1 .
To use Theorem 2.6, we need the second-order partial derivatives:
So
∂ 2 f
∂x 2=−6x, ∂ 2 f
∂y 2=−2, ∂ 2 f
∂y∂x = 1
D= ∂2 f
∂x 2(0,0)∂2 f
∂y 2(0,0)− ( ∂ 2 f
∂y∂x (0,0) ) 2=
(−6(0))(−2)−1 2 =−10
6
and ∂2 f( 1
∂x 2 6 , 1 ) (
12 =−1