Higher Engineering Mathematics, Sixth Edition

364 Higher Engineering Mathematics

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=0andx=2.

From equation (2),y=0.

Hence the stationary points are (0,0)

and (2, 0).

(iv)

∂^2 z

∂x^2

= 6 x− 6 ,

∂^2 z

∂y^2

=−8and

∂^2 z

∂x∂y

= 0

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

∂^2 z

∂x^2

=− 6 ,

∂^2 z

∂y^2

=−8and

∂^2 z

∂x∂y

= 0

For the point (2, 0),

∂^2 z

∂x^2

= 6 ,

∂^2 z

∂y^2

=−8and

∂^2 z

∂x∂y

= 0

0

z^52

1

z^52

2

z^52

4

z^5

2

MAX S

x

y

22

22

2

24 z 52

1

3

Figure 36.10

(vi)

(

∂^2 z

∂x∂y

) 2

=( 0 )^2 = 0

(vii) ( 0 , 0 )= 0 −(− 6 )(− 8 )=− 48

( 2 , 0 )= 0 −( 6 )(− 8 )= 48

(viii) Since ( 0 , 0 )<0and

(

∂^2 z

∂x^2

)

( 0 , 0 )

<0, the

point (0, 0) is a maximum pointand hencethe

maximum value is 0.

Since( 2 , 0 )>0,the point (2, 0) is a saddle

point.

The value of z at the saddle point is

23 − 3 ( 2 )^2 − 4 ( 0 )^2 + 2 =−2.

An approximate contour map representing the surface

f(x,y)is shown in Fig. 36.10 where a ‘holloweffect’ is

seen surroundingthe maximum point and a ‘cross-over’

occurs at the saddle pointS.

Problem 5. An open rectangular container is to

have a volume of 62.5m^3. Determine the least

surface area of material required.