VSLT (T&V)

EXAMPLE 9 :

Find 2z , 2z ,   z  for t
x2 y 2 y  x 

(b) z = e3xy + 4x 2 y

z xx = 9 y 2e3xy + 8 y, z xy = 9xye3xy + 3e3xy + 8x , z yy

the following functions .

= 9x 2e3xy 24



25

1.3 – Extremum Of Funct

COURSE FRAMEWORK AND ST

a) Determine the stationary point
equations up to third degree poly

b) Determine the nature of stationa
order partial derivatives test.

c) Solve problems related to econom

tions Of Two Variable

TUDENT LEARNING TIME (SLT)

(a, b) by solving simultaneous

ynomials.

ary point (a, b) by using the second

mics and business.

26

A necessary condition f

stationary point (critica

f (x, y): fx = 0

for the existence of a
al point) of a function

fy = 0

27

Discrim
(D-Test Fo

The type (nature) of
can be determined b
order partial deriva

D(x, y) = fxx (x, y)  f

minant
ormulae)

a stationary point
by using the second
ative test ( D test).

yy (x, y) − [ f xy (x, y)]2

28

D ( x, y) = fxx ( x, y) • f

D(x, y)  0

Min/max
exist

D(x, y)  0 S

D(x, y) = 0 Tes

f yy ( x, y ) −  fxy ( x, y )2

fxx (x, y)  0 maximum
exist

fxx (x, y)  0 minimum
exist

Saddle point exist

st fails

29

EXAMPLE 10: Find the statio

of the functio
and determine

Find stationary point :

1: Find fx , fy :

fx (x, y) = 2 y +

2: Let fx = 0, f y =

x=0

onary point (critical point )
on f (x, y) = 2xy + y2 + 2x2 +10

e its nature.

+ 4x f y (x, y) = 2x + 2 y

0. Solve simultaneously
0 y=0

30

3: Conclusion :Stat

(0,0) is a st

Determine its nature:

1: Find f xx , f xy , f yy

f xx = 4

2: Using D test form

D(x, y) = f xx (x, y)  f yy (x,

At point (0,0)

D(0,0) = 4  2 − [2]2 = 4  0
f xx = 4  0 (minimum exist)

te stationary point.
tationary point

f xy = 2 f yy = 2

mulae

, y) − [ f xy (x, y)]2

0 (min/max)

) (0,0) is a minimum point
31

EXAMPLE 11:

Find the maximum, minimu

of the functions. f (x, y) =

(0, 0) is a saddle point, (1,1) is minimum point

um or / and saddle point(s)

x3 + y 3 − 3xy

32



33

EXAMPLE 12:

A company produces two pro
units should be produced for
maximise the profit, if the pr

as P(x,y) = -3x2 + 4xy +100

much is the maximised profit

20 units product X and 5 units p

oducts, X and Y. How many
each product in order to
rofit functions is given
x-5y2-30 y +1500. How
t?

product Y.The maximise profit is RM2425.00 34



35



36

EXAMPLE 13:

The price function for producin

is p(x) = 200 − 1 x. The revenue

5

units of product Y is R(y) = 160

function of the two products is
Find

(a) the maximum profit to produ
and the quantity of each pro

(b) the price of product Y at ma

The maximise profit is RM14500[
The price of product y is RM135.

ng x units of product X is

e function for producing y

0 y − 1 y 2. The combine cost
4

C(x,y) = 100x + 70 y + 1 xy.

5

uce the two products
oduct,
aximum profit.

[200 units product x and 100 units product y] 37
.00



38



39

1.4 – Lagrange

COURSE FRAMEWORK AND S

a) Formulate the problem related to

subject to the constraint g ( x, y)
*Restrict g ( x, y) to linear only

b) Determine the critical points ( x, y

equations.
*Do not use the second order parti
maximum or minimum point.

c) Solve problems related to econom

e Multiplier

STUDENT LEARNING TIME (SLT)

maximize or minimize Z = f ( x, y)
)=0

y) by solving the simultaneous

ial derivative test (discriminant) to prove
mics and business.

40

Lagrange m

The below constrained
can be solved by using t
Method

Maximise/minim

Subject to g

multiplier

d optimization problem

the Lagrange Multiplier

mise Z = f (x, y)

g(x, y) = 0

41

STEP 1: Determine

F(x, y, ) = f (x,

STEP 2: Determine
by solving the simul

Fx (x, y, ) = 0 Fy (x,

STEP 3: Determine
maximum value(s)
value(s) into f (x, y)

e the new function.

, y) + [g(x, y) = 0]

Lagrange
Multiplier

the critical point (x, y, )

ltaneous equations.

y, ) = 0 F (x, y, ) = 0

e the minimum or / and
) by substituting the

42

EXAMPLE 14:

Maximise f (x, y) = −x2 −

g(x, y) = 2x + y −10.

Max value f (4,2) = 25

− y 2 + 45 subject to

43



44



45

EXAMPLE 15:

Minimise f (x, y) = x2 + 2 y

Min value f 10 ,− 5  = 59
 3 3 3

y 2 + 3 subject to x − y = 5.

46



47



48