Higher Engineering Mathematics, Sixth Edition

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

1. 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)]


2. 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

]

3. 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 )

⎤

⎦