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.