Higher Engineering Mathematics

MAXIMA, MINIMA AND SADDLE POINTS FOR FUNCTIONS OF TWO VARIABLES 359

G

(iv)

∂^2 z

∂x^2

= 6 x,

∂^2 z

∂y^2

= 6 yand

∂^2 z

∂x∂y

=

∂

∂x

(

∂z

∂y

)

=

∂

∂x

(− 6 x+ 3 y^2 )=− 6

(v) for (0, 0)

∂^2 z

∂x^2

=0,

∂^2 z

∂y^2

= 0

and

∂^2 z

∂x∂y

=− 6

for (2, 2),

∂^2 z

∂x^2

=12,

∂^2 z

∂y^2

= 12

and

∂^2 z

∂x∂y

=− 6

(vi) for (0, 0),

(

∂^2 z

∂x∂y

) 2

=(−6)^2 = 36

for (2, 2),

(

∂^2 z

∂x∂y

) 2

=(−6)^2 = 36

(vii) (0, 0)=

(

∂^2 z

∂x∂y

) 2

−

(

∂^2 z

∂x^2

)(

∂^2 z

∂y^2

)

= 36 −(0)(0)= 36

(2, 2)= 36 −(12)(12)=− 108

(viii) Since(0, 0)>0 then(0, 0) is a saddle point

Since(2, 2)<0 and

∂^2 z

∂x^2

>0, then(2, 2) is a

minimum point.

Now try the following exercise.

Exercise 144 Further problems on maxima,

minima and saddle points for functions of two

variables

1. Find the stationary point of the surface
   f(x,y)=x^2 +y^2 and determine its nature.
   Sketch the surface represented byz.
   [Minimum at (0, 0)]


2. Find the maxima, minima and saddle points
   for the following functions:
   (a)f(x,y)=x^2 +y^2 − 2 x+ 4 y+ 8
   (b)f(x,y)=x^2 −y^2 − 2 x+ 4 y+ 8

(c)f(x,y)= (^2) [x+ 2 y− 2 xy− 2 x^2 −y^2 + 4
(a) Minimum at (1,−2)
(b) Saddle point at (1, 2)
(c) Maximum at (0, 1)
]

3. Determine the stationary values of the func-
   tionf(x,y)=x^3 − 6 x^2 − 8 y^2 and distinguish
   between them. Sketch an approximate con-
   tour map to represent the surface[ f(x,y).
   Maximum point at (0, 0),
   saddle point at (4, 0)

]

4. Locate the stationary point of the function
   z= 12 x^2 + 6 xy+ 15 y^2.

[Minimum at (0, 0)]

5. Find the stationary points of the surface
   z=x^3 −xy+y^3 and distinguish between
   them. [
   saddle point at (0, 0),
   minimum at

( 1

3 ,

1

3

)

]

36.5 Further worked problems on

maxima, minima and saddle

points for functions of two

variables

Problem 3. Find the co-ordinates of the sta-

tionary points on the surface

z=(x^2 +y^2 )^2 −8(x^2 −y^2 )

and distinguish between them. Sketch the

approximate contour map associated withz.

Following the procedure:

(i)

∂z

∂x

=2(x^2 +y^2 )2x− 16 x and

∂z

∂y

=2(x^2 +y^2 )2y+ 16 y

(ii) for stationary points,

2(x^2 +y^2 )2x− 16 x= 0

i.e. 4 x^3 + 4 xy^2 − 16 x= 0 (1)

and 2(x^2 +y^2 )2y+ 16 y= 0

i.e. 4 y(x^2 +y^2 +4)= 0 (2)

(iii) From equation (1),y^2 =

16 x− 4 x^3

4 x

= 4 −x^2

Substitutingy^2 = 4 −x^2 in equation (2) gives

4 y(x^2 + 4 −x^2 +4)= 0

i.e. 32y=0 andy= 0