Higher Engineering Mathematics

MAXIMA, MINIMA AND SADDLE POINTS FOR FUNCTIONS OF TWO VARIABLES 361

G

h

− 2

j

− 4

z =^0

c

d

2 a e

S

g

f b − 2 x

y

4

i

2

z =^9

z = 128

Figure 36.9

Thus thez=9 contour also passes through (0, 1) and
(0,−1), shown asgandhin Fig. 36.9.

When, say,x=4 andy=0,

z=(4^2 )^2 −8(4^2 )=128.

whenz=128 andx=0, 128=y^4 + 8 y^2

i.e. y^4 + 8 y^2 − 128 = 0

i.e. (y^2 +16)(y^2 −8)= 0

from which,y=±

√
8 or complex roots.
Thus thez=128 contour passes through (0, 2.83)
and (0,−2.83), shown asiandjin Fig. 36.9.
In a similar manner many other points may be cal-
culated with the resulting approximate contour map
shown in Fig. 36.9. It is seen that two ‘hollows’
occur at the minimum points, and a ‘cross-over’
occurs at the saddle pointS, which is typical of such
contour maps.

Problem 4. Show that the function

f(x,y)=x^3 − 3 x^2 − 4 y^2 + 2

has one saddle point and one maximum point.

Determine the maximum value.

Letz=f(x,y)=x^3 − 3 x^2 − 4 y^2 +2.

Following the procedure:

(i)

∂z

∂x

= 3 x^2 − 6 xand

∂z

∂y

=− 8 y

(ii) for stationary points, 3x^2 − 6 x= 0 (1)

and − 8 y= 0 (2)

(iii) From equation (1), 3x(x−2)=0 from

which,x=0 andx=2.

From equation (2),y=0.