The Chemistry Maths Book, Second Edition

254 Chapter 9Functions of several variables

EXAMPLE 9.4Find the stationary points of the function

f(x, y) 1 = 1 x

3

1 + 16 xy

2

1 − 12 y

3

1 − 112 x

We have

and these are zero when

x

2

1 + 12 y

2

1 = 1 4, y(2x 1 − 1 y) 1 = 10

The second equation is satisfied when eithery 1 = 10 ory 1 = 12 x, and substitution of these

in the first equation gives the four stationary points

(2, 0), (−2, 0), (2 2 3, 4 2 3), (− 22 3, − 42 3)

0 Exercises 21–23

For a function of one variable, a stationary point atx 1 = 1 ais a local maximum if the

second derivative is negative,f′′(a) 1 < 10 , a local minimum iff′′(a) 1 > 10 , and may be a

point of inflection iff′′(a) 1 = 10. The corresponding conditions for a function of two

variables are

f

xx 1

< 1 0andf

yy

1 < 1 0 for a maximum (9.10a)

f

xx 1

1 0andf

yy 1

1 0 for a minimum (9.10b)

and

f

xx

f

yy

1 − 1 f

2

xy

1 > 1 0 for either a maximum or a minimum (9.10c)

If the quantity∆ 1 = 1 f

xx

f

yy

1 − 1 f

2

xy

is negative then the point is a saddle point; a maximum

in one direction and a minimum in another. If∆ 1 = 10 then further tests are required

to determine the nature of the point. The corresponding conditions are more

complicated for functions of more than two variables.

EXAMPLE 9.5The nature of the stationary points of Example 9.4.

The stationary points of the functionf(x, y) 1 = 1 x

3

1 + 16 xy

2

1 − 12 y

3

1 − 112 xare

(2, 0), (−2, 0), (2 2 3, 4 2 3), (− 22 3, − 42 3)

∂

∂

=+ −,

∂

∂

=−

f

x

xy

f

y

3 6 12 12 6xy y

22 2