Higher Engineering Mathematics, Sixth Edition

362 Higher Engineering Mathematics

Following the procedure:

(i)

∂z

∂x

= 2 (x^2 +y^2 ) 2 x− 16 x and

∂z

∂y

= 2 (x^2 +y^2 ) 2 y+ 16 y

(ii) for stationary points,

2 (x^2 +y^2 ) 2 x− 16 x= 0

i.e. 4x^3 + 4 xy^2 − 16 x=0(1)

and 2(x^2 +y^2 ) 2 y+ 16 y= 0

i.e. 4 y(x^2 +y^2 + 4 )=0(2)

(iii) From equation (1),y^2 =

16 x− 4 x^3

4 x

= 4 −x^2

Substitutingy^2 = 4 −x^2 in equation (2) gives

4 y(x^2 + 4 −x^2 + 4 )= 0

i.e. 32y=0andy= 0

Wheny=0 in equation (1), 4x^3 − 16 x= 0

i.e. 4 x(x^2 − 4 )= 0

from which,x=0orx=± 2

The co-ordinates of the stationary points are

(0, 0), (2, 0) and (−2, 0).

(iv)

∂^2 z

∂x^2

= 12 x^2 + 4 y^2 − 16 ,

∂^2 z

∂y^2

= 4 x^2 + 12 y^2 +16 and

∂^2 z

∂x∂y

= 8 xy

(v) For the point (0, 0),

∂^2 z

∂x^2

=− 16 ,

∂^2 z

∂y^2

=16 and

∂^2 z

∂x∂y

= 0

For the point (2, 0),

∂^2 z

∂x^2

= 32 ,

∂^2 z

∂y^2

=32 and

∂^2 z

∂x∂y

= 0

For the point (−2, 0),

∂^2 z

∂x^2

= 32 ,

∂^2 z

∂y^2

=32 and

∂^2 z

∂x∂y

= 0

(vi)

(

∂^2 z

∂x∂y

) 2

=0 for each stationary point

(vii) ( 0 , 0 ) =( 0 )^2 −(− 16 )( 16 )= 256

( 2 , 0 ) =( 0 )^2 −( 32 )( 32 )=− 1024

(− 2 , 0 )=( 0 )^2 −( 32 )( 32 )=− 1024

(viii) Since( 0 , 0 )>0,the point (0, 0) is a saddle

point.

Since( 0 , 0 )<0and

(

∂^2 z

∂x^2

)

( 2 , 0 )

>0,the point

(2, 0) is a minimum point.

Since (− 2 , 0 )<0and

(

∂^2 z

∂x^2

)

(− 2 , 0 )

>0, the

point (−2, 0) is a minimum point.

Looking down towards thex-yplane from above, an

approximate contour map can be constructed to repre-

sent the value ofz. Such a map is shown in Fig. 36.9.

To produce a contour map requires a large number of

x-yco-ordinates to be chosen and the values ofzat

each co-ordinate calculated. Here are a few examples of

points used to construct the contour map.

Whenz= 0 , 0 =(x^2 +y^2 )^2 − 8 (x^2 −y)^2

In addition, when, say,y=0 (i.e. on thex-axis)

0 =x^4 − 8 x^2 ,i.e.x^2 (x^2 − 8 )= 0

from which, x=0orx=±

√

8

Hencethecontourz=0 crossesthex-axisat 0and±

√

8,

i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown

as points,S,aandbrespectively.

Whenz=0andx= 2 then

0 =( 4 +y^2 )^2 − 8 ( 4 −y^2 )

i.e. 0= 16 + 8 y^2 +y^4 − 32 + 8 y^2

i.e. 0=y^4 + 16 y^2 − 16

Let y^2 =p,thenp^2 + 16 p− 16 =0and

p=

− 16 ±

√

162 − 4 ( 1 )(− 16 )

2

=

− 16 ± 17. 89

2

= 0 .945 or− 16. 945

Hence y=

√

p=

√

( 0. 945 )or

√

(− 16. 945 )

=± 0 .97 or complex roots.