Maxima, minima and saddle points for functions of two variables 365
z
x
y
Figure 36.11
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
Substituting in equation (2) gives:
S=xy+ 2 y
(
62. 5
xy
)
+ 2 x
(
62. 5
xy
)
i.e. S=xy+
125
x
+
125
y
which is a function of two variables
∂s
∂x
=y−
125
x^2
=0 for a stationary point,
hence x^2 y= 125 ( 3 )
∂s
∂y
=x−
125
y^2
=0 for a stationary point,
hence xy^2 = 125 ( 4 )
Dividing equation (3) by (4) gives:
x^2 y
xy^2
= 1 ,i.e.
x
y
= 1 ,i.e.x=y
Substitutingy=xin equation (3) givesx^3 =125, from
which,x=5m.
Hencey=5malso
From equation (1), (5) (5)z= 62. 5
from which, z=
62. 5
25
= 2 .5m
∂^2 S
∂x^2
=
250
x^3
,
∂^2 S
∂y^2
=
250
y^3
and
∂^2 S
∂x∂y
= 1
Whenx=y= 5 ,
∂^2 S
∂x^2
= 2 ,
∂^2 S
∂y^2
=2and
∂^2 S
∂x∂y
= 1
=( 1 )^2 −( 2 )( 2 )=− 3
Since<0and
∂^2 S
∂x^2
>0, then the surface areaSis a
minimum.
Hence the minimum dimensions of the container to have
a volume of 62.5m^3 are5mby5mby2.5m.
From equation (2),minimum surface area,S
=( 5 )( 5 )+ 2 ( 5 )( 2. 5 )+ 2 ( 5 )( 2. 5 )
=75m^2
Now try the following exercise
Exercise 144 Further problemson
maxima,minima and saddle points for
functions of two variables
- The functionz=x^2 +y^2 +xy+ 4 x− 4 y+ 3
has one stationary value. Determine its
co-ordinates and its nature.
[Minimum at (−4, 4)] - An open rectangular container is to have a vol-
ume of 32m^3. Determine the dimensions and
the total surface area such that the total surface
area is a minimum.
[
4mby4mby2m,
surface area=48m^2
]
- Determine the stationary values of the
function
f(x,y)=x^4 + 4 x^2 y^2 − 2 x^2 + 2 y^2 − 1
and distinguish between them.
⎡
⎣
Minimum at( 1 , 0 ),
minimum at(− 1 , 0 ),
saddle point at( 0 , 0 )
⎤
⎦