362 DIFFERENTIAL CALCULUS
Hence the stationary points are (0, 0)
and (2, 0).(iv)∂^2 z
∂x^2= 6 x−6,∂^2 z
∂y^2=−8 and∂^2 z
∂x∂y= 0(v) For the point (0, 0),∂^2 z
∂x^2=−6,∂^2 z
∂y^2=−8 and∂^2 z
∂x∂y= 0For the point (2, 0),∂^2 z
∂x^2=6,∂^2 z
∂y^2=−8 and∂^2 z
∂x∂y= 0(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)<0 and(
∂^2 z
∂x^2)(0, 0)<0,thepoint (0, 0) is a maximum pointand hence
the maximum value is 0.y20− 22S− (^24) x
z =
2
z^ =^
−^1
z^ =^
−^2
z^ =^ −^4
z (^) =
(^) − 1
MAX
Figure 36.10
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
‘hollow effect’ is seen surrounding the maximum
point and a ‘cross-over’ occurs at the saddle
pointS.
Problem 5. An open rectangular container is to
have a volume of 62.5 m^3. Determine the least
surface area of material required.
Let the dimensions of the container bex,yandzas
shown in Fig. 36.11.
Volume V=xyz= 62. 5 (1)
Surface area, S=xy+ 2 yz+ 2 xz (2)
From equation (1),z=
62. 5
xy