358 DIFFERENTIAL CALCULUS

y

1

(^2) z = 4 z = 9 z = 16
12 x
z = 1
Figure 36.8
Problem 2. Find the stationary points of the
surface f(x,y)=x^3 − 6 xy+y^3 and determine
their nature.
Letz=f(x,y)=x^3 − 6 xy+y^3
Following the procedure:
(i)
∂z
∂x
= 3 x^2 − 6 yand
∂z
∂y
=− 6 x+ 3 y^2
(ii) for stationary points, 3x^2 − 6 y= 0 (1)
and − 6 x+ 3 y^2 = 0 (2)
(iii) from equation (1), 3x^2 = 6 y
and y=
3 x^2
6

1
2
x^2
and substituting in equation (2) gives:
− 6 x+ 3
(
1
2
x^2
) 2
= 0
− 6 x+
3
4
x^4 = 0
3 x
(
x^3
4
− 2
)
= 0
from which,x=0or
x^3
4
− 2 = 0
i.e.x^3 =8 andx= 2
Whenx=0,y=0 and whenx=2,y=2 from
equations (1) and (2).
Thus stationary points occur at (0, 0)
and (2, 2).