18MAB101T- Calculus And Linear Algebra

Unit II-Functions of Several Variables

Dr. S. SABARINATHAN,
Assistant Professor,
Department of Mathematics,
Kattankulathur-603 203.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR

SRM ()Institute of Science and Technology 1 / 66
 Maxima and Minima of Functions two variables

Maximum Value: A function f (x, y) is said to have a maximum value at

x = a, y = b if f (a, b) > f (a + h, b + k ), for small and independent values
of h and k , positive or negative.

Minimum Value: A function f (x, y ) is said to have a maximum value at

x = a, y = b if f (a, b) < f (a + h, b + k ), for small and independent values
of h and k , positive or negative.

Extreme Value: f (a, b) is said to be an extremum value of f (x, y ) if it is

either maximum or minimum.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 2 / 66

 Working rule to find extreme values (Necessary
Conditions)
∂f ∂f
Step 1: Find and .
∂x ∂y
∂f ∂f
Step 2: Solve the equations = 0 and = 0 simultaneously.
∂x ∂y
Let the solutions be (a, b), (c, d), . . .
∂f ∂f
Stationary Points: The point (a, b) at which = 0 and = 0 are
∂x ∂y
called stationary points of the function f (x, y).
Stationary values: The values of f (x, y ) at the stationary points are
called stationary values of the function f (x, y ).
Note: Every extremum value is a stationary value but a stationary value
need not be an extremum.
Notations:
∂f ∂f ∂ 2f ∂ 2f ∂ 2f
p= ,q = ,r = , s = and t =
∂x ∂y ∂ x2 ∂ x∂ y ∂y2
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 3 / 66
 Sufficient Condition for Maxima and Minima

Let (a, b) be a stationary point. Then if

rt − s2 > 0 at (a, b) and r < 0 (t < 0) then f (a, b) is maximum value.
rt − s2 > 0 at (a, b) and r > 0 (t > 0) then f (a, b) is minimum value.
rt − s2 < 0 at (a, b) then f (a, b) has neither a maximum nor a mini-
mum value. In this case, the point (a, b) is called a saddle point of
the function f (x, y ).
if rt − s2 = 0, then the case is doubtful and hence further investiga-
tions are required.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 4 / 66

 Example 1:

Discuss the maximum and minimum of x 2 + y 2 + 6x + 12.

Solution: Let f (x, y ) = x 2 + y 2 + 6x + 12

Now p = 2x + 6, q = 2y, r = 2, s = 0 and t = 2
The stationary points are given by p = 0, q = 0
⇒ 2x + 6 = 0 and 2y = 0
⇒ x = −3 and y = 0
. ˙ . (-3, 0) is the stationary point.
(-3,0)
r 2 (> 0)
s 0
t 2 (> 0)
rt − s2 4 (> 0)
Hence f (x, y) is minimum when x = −3 and y = 0.
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 5 / 66
 Example 2:

Examine f (x, y) = x 3 + y 3 − 3xy for maximum and minimum values.

Solution: Let f (x, y ) = x 3 + y 3 − 3xy

Now p = 3x 2 − 3y, q = 3y 2 − 3x, r = 6x, s = −3 and t = 6y
The stationary points are given by p = 0, q = 0

⇒ 3x 2 − 3y = 0 and 3y 2 − 3x = 0
x2 = y (1)
2
y =x and (2)
√
Substituting (2) in (1), we get x 2 = x

⇒ x 4 = x ⇒ x(x 3 − 1) = 0
⇒ x = 0, 1
. ˙ . y = 0, 1
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 6 / 66
 Example 2: (Contd.)

Therefore (0, 0) and (1, 1) are the stationary points.

(0,0) (1,1)
r = 6x 0 6 (> 0)
s = −3 -3 -3
t = 6y 0 6 (> 0)
rt − s2 -9 (<0) 27 (>0)

At (0,0) is a saddle point and at (1,1) is a point of minimum value.

. ˙ . the minimum value of f (1, 1) = 1 + 1 − 3 = −1.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 7 / 66

 Example 3:

Find the maximum or minimum value of sin x + sin y + sin(x + y ).

Solution: Given f (x, y ) = sin x + sin y + sin(x + y)

Now p = cos x + cos(x + y), q = cos y + cos(x + y)
r = − sin x − sin(x + y ), t = − sin y − sin(x + y) and s = − sin(x + y)
The stationary points are obtained by equating p = 0 and q = 0
⇒ cos x + cos(x + y) = 0 and cos y + cos(x + y) = 0
. ˙ . cos x = − cos(x + y )
⇒ cos x = cos(π − (x + y)) ⇒ x = π − (x + y)
2x + y = π (3)
Similarly q = 0, we get
x + 2y = π (4)
π π
Solving (3) and (4), we get x = and y =
3
π π  3
. ˙ . the stationary points is , .
3 3
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 8 / 66
 Example 3: (contd.)

π π 
,
√3 3
r = − sin x − sin(x + y) − 3√(< 0)
3
s = − sin(x + y ) −
√ 2
t = − sin y − sin(x + y) − 3 (< 0)
9
rt − s2 (> 0)
4
π π 
. ˙ . the point
, is a maximum point.
3 3
Hence the maximum value of
√
π π  π π 2π 3 3
f , = sin + sin + sin = .
3 3 3 3 3 2

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 9 / 66

 Example 4:

Find the extreme values of the function

f (x, y ) = x 3 + y 3 − 3x − 12y + 20.

Solution: Given f (x, y ) = x 3 + y 3 − 3x − 12y + 20

Now p = 3x 2 − 3, q = 3y 2 − 12, r = 6x, s = 0 and t = 6y
The stationary points are obtained by equating p = 0 and q = 0

p=0 q=0
⇒ 3x 2 − 3 = 0 ⇒ 3y 2 − 12
⇒ x2 − 1 = 0 ⇒ y2 − 4 = 0
⇒ x = ±1 ⇒ y = ±2

. ˙ . the stationary points are (1,2), (1,-2), (-1,2), (-1,-2).

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 10 / 66

 Example 4: (Contd.)

(1,2) (1,-2) (-1,2) (-1,-2)

r = 6x 6 (> 0) 6 (> 0) -6 (< 0) -6 (> 0)
s=0 0 0 0 0
t = 6y 12 (> 0) -12 (< 0) 12 (> 0) -12 (< 0)
rt − s2 72 (<0) -72 (<0) -72 (< 0) 72 (> 0)
min. saddle saddle max.

Hence the maximum value of f (−1, −2) is 38 and the minimum value
of f (1, 2) is 2.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 11 / 66

 Example 4:

a3 a3
Examine for extreme values of f (x, y) = xy + + .
x y
a3 a3
Solution: Given f (x, y ) = xy + +
x y
a3 a3 2a3 2a3
Now p = y − 2 , q = x − 2 , r = 3 , s = 1 and t = 3
x y x y
The stationary points are obtained by equating p = 0 and q = 0

a3
⇒y− =0 and (5)
x2
a3
x− =0 (6)
y2
a3
From (5) ⇒ y =
x2
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 12 / 66
 Example 4: (Contd.)

Substituting this value in (6), we get

x4
x−=0
a3
x3
 
⇒ x 1− 3 = 0
a
⇒ x = 0, a.
When x = 0 ⇒ y = ∞ and When x = a ⇒ y = a
Omit (0, ∞), the stationary point is (a, a).

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 13 / 66

 Example 4: (Contd.)

(a, a)
2a3
r= 2 (> 0)
x3
s=1 1
2a3
t= 3 2 (> 0)
y
rt − s2 3 (> 0)

. ˙ . the point (a, a) is a minimum point.

Hence the minimum value of f (a, a) = 3a2 .

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 14 / 66

 Example 5:

Examine f (x, y) = x 3 + y 3 − 3axy for maxima and minima.

Solution: Given f (x, y ) = x 3 + y 3 − 3axy

Now p = 3x 2 − 3ay, q = 3y 2 − 3ax, r = 6x, s = −3a and t = 6y
The stationary points are obtained by equating p = 0 and q = 0

⇒ 3x 2 − 3ay = 0 and 3y 2 − 3ax = 0

⇒ x 2 = ay and y 2 = ax
Solving these two equations, we get (0,0) and (a, a).
Therefore the stationary points are (0,0) and (a, a).

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 15 / 66

 Example 5: (Contd.)

(0,0) (a, a)
r = 6x 0 6a
s = −3a -3a -3a
t = 6y 0 6a
rt − s2 -9 (<0) 27a2 (>0)

Hence the point (a, a) is a minimum if a > 0 and (a, a) is a maximum if

a < 0.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 16 / 66

 Example 6:

Find the extreme values of f (x, y) = x 3 y 2 (1 − x − y).

Solution: Given f (x, y ) = x 3 y 2 (1 − x − y ) = x 3 y 2 − x 4 y 2 − x 3 y 3

Now p = 3x 2 y 2 − 4x 3 y 2 − 3x 2 y 2 = x 2 y 2 (3 − 4x − 3y),
q = 2x 3 y − 2x 4 y − 3x 3 y 2 = x 3 y (2 − 2x − 3y ),
r = 6xy 2 − 12x 2 y 2 − 6xy 3 = 6xy 2 (1 − 2x − y ),
s = 6x 2 y − 8x 3 y − 9x 2 y 2 = x 2 y(6 − 8x − 9y ) and
t = 2x 3 − 2x 4 − 6x 3 y = x 3 (2 − 2x − 6y )
The stationary points are obtained by equating p = 0 and q = 0

⇒ x 2 y 2 (3 − 4x − 3y ) = 0 and x 3 y (2 − 2x − 3y ) = 0

⇒ x = 0, y = 0, 4x + 3y = 3 and x = 0, y = 0, 2x + 3y = 2

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 17 / 66

 Example 6: (Contd.)

⇒ 4x + 3y = 3 and (7)
2x + 3y = 2 (8)
 
1 1
Solving these two equations, we get , .
2 3
Put x = 0 in (7), we get y = 1
3
Put y = 0 in (7), we get x =
4
2
Put x = 0 in (8), we get y =
3
Put y = 0 in (8), we get x = 1      
1 1 2 3
. ˙ . the stationary points are (0,0), , , (0,1), 0, , , 0 and
2 3 3 4
(1,0).

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 18 / 66

 Example 6: (Contd.)

     
1 1 2 3
(0,0) , (0,1) 0, ,0 (1,0)
2 3 3 4
1
r = 6xy 2 (1 − 2x − y ) 0 − (< 0) 0 0 0 0
9
1
s = x 2 y (6 − 8x − 9y) 0 − 0 0 0 0
12
1 27
t = x 3 (2 − 2x − 6y ) 0 − (< 0) 0 0 0
8 128
1
rt − s2 0 (> 0) 0 0 0 0
144
inco. Max. incon. inco. inco. inco.
 
1 1
Therefore , is a maximum point.
2 3
Hence the
 maximum
 3  value
  of
1 2
 
1 1 1 1 1 1
f , = 1− − = .
2 3 2 3 2 3 432
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 19 / 66
 Lagrange’s Method of Undetermined Multipliers
This method is to find the maximum or minimum value of a function of
three or more variables, given the constraints.
Let
u = f (x, y, z) (9)
be a function of three variables which is to be tested for maximum or
minimum value, subject to the condition (constraint)

g(x, y, z) = 0 (10)

By total differentials, we have

∂f ∂f ∂f
du = . dx + . dy + . dz by (9) (11)
∂x ∂y ∂z

∂g ∂g ∂g
0= . dx + . dy + . dz by (10) (12)
∂x ∂y ∂z
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 20 / 66
 Lagrange’s Method of Undetermined Multipliers
The conditions for f (x, y, z) to have a maximum point or a minimum
point is du = 0. Therefore (11), we get

∂f ∂f ∂f
. dx + . dy + . dz = 0 (13)
∂x ∂y ∂z

Multiply (12) by λ , we get

∂g ∂g ∂g
λ . dx + λ . dy + λ . dz = 0 (14)
∂x ∂y ∂z

Adding (13) and (14), we get

     
∂f ∂g ∂f ∂g ∂f ∂g
+λ dx + +λ dy + +λ dz = 0
∂x ∂x ∂y ∂y ∂z ∂z

Here λ is called the Lagrange multiplier.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 21 / 66
 Lagrange’s Method of Undetermined Multipliers

Now we shall choose λ such that

∂f ∂g
+λ =0
∂x ∂x
∂f ∂g
+λ =0
∂y ∂y
∂f ∂g
+λ =0
∂z ∂z
Solving the above equations along with the given relation, we get the
values of x, y , z and λ .
These values give finally the required maximum or minimum value of
the function f (x, y, z).

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 22 / 66

 Working Rule

To find the maximum and minimum values of f (x, y, z) where x, y, z are

subject to the constraint g(x, y , z)
We define a Function

F (x, y, z) = f (x, y , z) + λ g(x, y , z)

∂F ∂F ∂F
Find , and
∂x ∂y ∂z
∂F ∂F ∂F
Set = 0, = 0 and = 0 and then solve we get x, y , z.
∂x ∂y ∂z

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 23 / 66

 Example 1:

A rectangular box open at the top is to have volume of 32 cubic ft. Find
the dimensions in order that the total surface area is minimum.
Solution: Given
g(x, y, z) = xyz − 32 = 0 (15)
Let x, y, z be the dimension of rectangular box open at the top.
Total surface area (S): f (x, y , z) = xy + 2xz + 2yz
We define the function
F (x, y, z) = xy + 2xz + 2yz + λ (xyz − 32)
At the critical points, we have
∂f ∂g
+λ = 0 ⇒ y + 2z + λ yz = 0 (16)
∂x ∂x
∂f ∂g
+λ = 0 ⇒ x + 2z + λ xz = 0 (17)
∂y ∂y
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 24 / 66
 Example 1: (Contd.)

∂f ∂g
+λ = 0 ⇒ 2x + 2y + λ xy = 0 (18)
∂z ∂z
(16) × x − (17) × y ⇒ 2(zx − zy) = 0 ⇒ z 6= 0, x − y = 0

⇒x =y (19)

⇒ y 2 − 2yz = 0 by (19)
⇒ y(y − 2z) = 0 ⇒ y =
6 0, y − 2z = 0
y
⇒z = (20)
2
Using (19) and (20) in (16) we get x = 4.
. ˙ . y = 4, z = 2.
Hence the dimensions are 4cm, 4cm and 2cm.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 25 / 66

 Example 2:
Find the volume of the largest rectangular parallelopiped that can be
x 2 y 2 z2
inscribed in the ellipsoid 2 + 2 + 2 = 1.
a b c
Solution: The given ellipsoid is
x 2 y 2 z2
g(x, y , z) =
+ + −1 = 0 (21)
a2 b 2 c 2
Let 2x, 2y, 2z be the dimensions of the required parallelopiped.
The volume of the parallelopiped (V ) : f (x, y, z) = 8xyz
We define the function
 2
y 2 z2

x
F (x, y , z) = 8xyz + λ + + −1
a2 b 2 c 2
At the critical points, we have
 
∂f ∂g 2x
+λ = 0 ⇒ 8yz + λ =0 (22)
∂x ∂x a2
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 26 / 66
 Example 2: (Contd.)
 
∂f ∂g 2y
+λ = 0 ⇒ 8xz + λ =0 (23)
∂y ∂y b2
 
∂f ∂g 2z
+λ = 0 ⇒ 8xy + λ =0 (24)
∂z ∂z c2
(22) × x + (23) × y + (24) × z, we get
 2
y 2 z2

x
24xyz + 2λ + + =0
a2 b 2 c 2
⇒ 2λ = −24xyz by (21)
λ = −12xyz (25)
Using (25) in (24), we get
 
2z
8xy + (−12xyz) =0
c2
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 27 / 66
 Example 2: (Contd.)

3z 2
 
⇒ 8xy 1 − 2 = 0
c
3z 2
⇒ =1
c2
c
⇒ z = √ , since x 6= 0, y 6= 0
3
b a
Similarly y = √ , c = √
3 3
8abc
Hence the volume of rectangular parallelopiped is V = √ units.
3 3

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 28 / 66

 Example 3:

Find the dimensions of the rectangular box, open at the top of

maximum capacity whose surface is 432 sq.cm.

Solution: Let x, y, z be the dimensions of the rectangular bx, open at

the top.
Given its surface area
g(x, y , x) = xy + 2yz + 2zx − 432 = 0 (26)
The volume is (V ) : f (x, y, z) = xyz
We define the function
F (x, y, z) = xyz + λ (xy + 2yz + 2zx − 432)
At the critical points, we get
yz + λ (y + 2z) = 0 (27)
xz + λ (x + 2z) = 0 (28)
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 29 / 66
 Example 3: (Contd.)

xy + λ (2y + 2x) = 0 (29)

(27) × x − (28) × y ⇒ 2λ z(x − y) = 0

⇒ x = y since λ 6= 0, z 6= 0 (30)

(28) × x − (29) × z ⇒ λ y(x − 2z) = 0

x
⇒z = since λ 6= 0, y 6= 0 (31)
2
Using (30) and (31) in (26), we get x = 12.
. ˙ . y = 12, z = 6.
Hence the dimensions of the rectangular box are 12 cm, 12 cm and 6
cm.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 30 / 66

 Example 4:

Find the maximum and minimum distance of the point (3,4,12) from the
sphere x 2 + y 2 + z 2 = 1.

Solution: Let (x, y , z) be any point on the sphere.

Given
g(x, y, z) = x 2 + y 2 + z 2 − 1 = 0 (32)
Distance
p of the point (x, y , z) from (3, 4, 12) is given by
d = (x − 3)2 + (y − 4)2 + (z − 1)2
f (x, y , z) = square of the distance from the point (3, 4, 12) to the sphere

i.e. f (x, y , z) = (x − 3)2 + (y − 4)2 + (z − 1)2

We define the function
 
F (x, y , z) = (x − 3)2 + (y − 4)2 + (z − 1)2 + λ x 2 + y 2 + z 2 − 1

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 31 / 66

 Example 4: (Contd.)

At the critical points, we have

2(x − 3) + 2λ x = 0 (33)

2(y − 4) + 2λ y = 0 (34)
2(z − 12) + 2λ z = 0 (35)
From (33), (34) and (35), we get

3 4 12
x= ,y= ,z = (36)
1+λ 1+λ 1+λ
Using (36) in (32), we get

(1 + λ )2 = 169 ⇒ 1 + λ = ±13 (37)

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 32 / 66

 Example 4: (Contd.)

Using (37) in (36), we get

   
3 4 12 −3 −4 −12
, , and , ,
13 13 13 13 13 13

Therefore
s the distance are
3 2 4 2 12 2
    
3− + 4− + 12 − = 12 and
13 13 13
s
3 2 4 2 12 2
    
3+ + 4+ + 12 + = 14
13 13 13
Hence the maximum distance is 14 and the minimum distance is 12.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 33 / 66

 Example 5:
3 4 5
If + + = 6 find the values of x, y, z which make x + y + z is
x y z
minimum.
Solution:
Given
3 4 5
+ + −6 = 0
g(x, y, z) = (38)
x y z
The required function is f (x, y , z) = x + y + z
We define the function
 
3 4 5
F (x, y, z) = x + y + z + λ + + −6
x y z
At the critical points, we have
3λ x2
1− = 0 ⇒ λ = (39)
x2 3
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 34 / 66
 Example 5: (Contd.)

4λ y2
1− = 0 ⇒ λ = (40)
y2 4
5λ z2
= 0 ⇒ λ 1−
= (41)
z2 5
From (39), (40) and (41), we get
x 2 y 2 z2
λ= = =
3 4 5
√ √ √
⇒ x = 3λ , y = 2 λ , z = 5λ (42)
Using (42) in (38), we get
3 4 5
√ +√ +√ =6
3λ λ 5λ
√ √
√ 3+2+ 5
⇒ λ=
6
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 35 / 66
 Example 5: (Contd.)

√
Substituting λ in (42), we get
√
3 √ √
x= ( 3 + 2 + 5)
6
1 √ √
y = ( 3 + 2 + 5)
6
√
5 √ √
z= ( 3 + 2 + 5)
6

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 36 / 66

 Example 6:

Find the maximum value of x m y n z p when x + y + z = a.

Solution: Given
g(x, y, z) = x + y + z − a = 0 (43)
The required function is f (x, y , z) = x m y n z p
We define the function

F (x, y, z) = x m y n z p + λ (x + y + z − a)

At the critical points, we have

mx m−1 y n z p + λ = 0 (44)

nx m y n−1 z p + λ = 0 (45)
px m y n z p−1 + λ = 0 (46)
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 37 / 66
 Example 6: (Contd.)

From equations (44), (45) and (46), we get

−λ = mx m−1 y n z p = nx m y n−1 z p = px m y n z p−1

m n p
= = ⇒
x y z
m+n+p
=
x +y +z
m+n+p
= by (43)
a
am an ap
.˙. x = ,y= , z=
m+n+p m+n+p m+n+p
am+n+p mm nn pp
The maximum value of f = .
(m + n + p)m+n+p

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 38 / 66

 Jacobians

If u and v are functions of the two independent variables x and y , then

the determinant
∂u ∂u
∂x ∂y
∂v ∂v
∂x ∂y
iscalledthe Jacobian of u, v with respect to x, y . It is denoted by
u, v ∂ (u, v )
J or .
x, y ∂ (x, y)

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 39 / 66

 Jacobians: Note

The Jacobian of u, v , w with respect to x, y , z is

∂u ∂u ∂u
∂x ∂y ∂z
∂ (u, v , w) ∂v ∂v ∂v
=
∂ (x, y, z) ∂x ∂y ∂z
∂w ∂w ∂w
∂x ∂y ∂z

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 40 / 66

 Properties of Jacobians

Property 1:
If J1 is the Jacobian of u, v with respect to x, y and J2 is the Jacobian
of x, y with respect to u, v then J1 J2 = 1.
∂ (u, v ) ∂ (x, y )
i.e. × = 1.
∂ (x, y) ∂ (u, v )

Proof:
∂u ∂u ∂x ∂x
∂ (u, v ) ∂ (x, y)
× = ∂∂ vx ∂y × ∂u
∂v ∂y
∂v
∂y
∂ (x, y) ∂ (u, v )
∂x ∂y ∂u ∂v
∂u ∂x ∂u ∂y ∂u ∂x ∂u ∂y
. + . . + .
= ∂∂ vx ∂∂ u ∂y ∂x
x ∂v ∂y
∂x ∂v ∂y ∂v
∂v ∂x ∂v ∂y (47)
. + . . + .
∂x ∂u ∂y ∂u ∂x ∂v ∂y ∂v

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 41 / 66

 Property 1: (Contd.)
Let u = u(x, y) and v = v (x, y)
Differentiating partially w.r.to u and v , we get
∂u ∂u ∂x ∂u ∂y 

=1= . + . 
∂u ∂x ∂u ∂y ∂u 



∂u ∂u ∂x ∂u ∂y 


=0= . + .  
∂v ∂x ∂v ∂y ∂v

(48)
∂v ∂v ∂x ∂v ∂y 

=0= . + . 
∂u ∂x ∂u ∂y ∂u


∂v ∂v ∂x ∂v ∂y 


=1= . + .  
∂v ∂x ∂v ∂y ∂v
Substituting (48) in (47), we get
∂ (u, v ) ∂ (x, y ) 1 0
× = =1
∂ (x, y) ∂ (u, v ) 0 1

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 42 / 66

 Properties of Jacobians
Property 2:
If u, v are functions of r , s where r , s are functions of x, y then

∂ (u, v ) ∂ (u, v ) ∂ (r , s)
= ×
∂ (x, y) ∂ (r , s) ∂ (x, y )

Proof: R.H.S
∂u ∂u ∂r ∂r
∂ (u, v ) ∂ (r , s)
× = ∂∂ vr ∂∂ vs × ∂∂ xs ∂y
∂s
∂ (r , s) ∂ (x, y )
∂r ∂s ∂x ∂y
∂u ∂r ∂u ∂s ∂u ∂r ∂u ∂s
. + . . + .
= ∂∂ vr ∂∂ xr ∂∂ vs ∂∂ xs ∂r ∂y ∂s ∂y
∂v ∂r ∂v ∂s
. + . . + .
∂r ∂x ∂s ∂x ∂r ∂y ∂s ∂y

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 43 / 66

 Property 1: (Contd.)

∂u ∂u
∂ (u, v ) ∂ (r , s)
× = ∂∂ vx ∂∂ vy
∂ (r , s) ∂ (x, y)
∂x ∂y
∂ (u, v )
= = L.H.S
∂ (x, y)
Note:
∂ (u, v , w) ∂ (u, v , q) ∂ (r , s, t)
1 = ×
∂ (x, y, z) ∂ (r , s, t) ∂ (x, y , z)
2 If u, v , w are functionally dependent functions of three independent
∂ (u, v , w)
variables x, y , z then =0
∂ (x, y, z)
3 If u, v , w are said to be functionally dependent, if each can be ex-
pressed interms of the others.
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 44 / 66
 Example 1:

If x = u 2 − v 2 and y = 2uv , find the Jacobian of x and y with respect to

u and v .
Solution:
∂x ∂x
∂ (x, y )
= ∂u ∂v
∂ (u, v ) ∂y ∂y
∂u ∂v
2u −2v
=
2v 2u
= 4(u 2 + v 2 )

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 45 / 66

 Example 2:

If x = r sin θ cos φ , y = r sin θ sin φ , z = r cos θ , show that

∂ (x, y , z)
= r 2 sin θ .
∂ (r , θ , φ )

Proof:
∂x ∂x ∂x
∂r ∂θ ∂φ
∂ (x, y, z) ∂y ∂y ∂y
=
∂ (r , θ , φ ) ∂r ∂θ ∂φ
∂z ∂z ∂z
∂r ∂θ ∂φ
sin θ cos φ r cos θ cos φ −r sin θ sin φ
= sin θ sin φ r cos θ sin φ r sin θ cos φ
cos θ −r sin θ 0

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 46 / 66

 Example 2: (Contd.)

sin θ cos φ cos θ cos φ − sin φ

∂ (x, y, z) 2
= r sin θ sin θ sin φ cos θ sin φ cos φ
∂ (r , θ , φ )
cos θ − sin θ 0
(
cos θ cos φ − sin φ
= r 2 sin θ cos θ
cos θ sin φ cos φ
)
sin θ cos φ − sin φ
+ sin θ by expanding the third row
sin θ sin φ cos φ
"
 
= r sin θ cos θ cos θ cos2 φ + cos θ sin2 φ
2

#
 
2 2
+ sin θ sin θ cos φ + sin θ sin φ = r 2 sin θ

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 47 / 66

 Example 3:
yz zx xy
If u = ,v = ,w = . Show that the Jacobian of u, v , w with
x y z
respect to x, y, z is 4.

Proof:
∂u ∂u ∂u
∂x ∂y ∂z
∂ (u, v , w) ∂v ∂v ∂v
=
∂ (x, y, z) ∂x ∂y ∂z
∂w ∂w ∂w
∂x ∂y ∂z
yz z y
− 2
x x x
z zx x
= y − 2
y y
y x xy
− 2
z z z
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 48 / 66
 Example 3: (Contd.)

−yz zx yx
∂ (u, v , w) 1
= 2 2 2 zy −zx xy
∂ (x, y , z) x y z
yz xz −xy
−1 1 1
x 2y 2z 2
= 1 −1 1
x 2y 2z 2
1 1 −1
= −1(1 − 1) − 1(−1 − 1) + 1(1 + 1) = 4

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 49 / 66

 Example 4:
√
Are the functions u = sin−1 x + sin−1 y and v = x
p
1 − y 2 + y 1 − x 2,
functionally dependent? (Given x 2 < 1, y 2 < 1.)

Solution:
∂u ∂u
∂ (u, v )
= ∂∂ vx ∂y
∂v
∂ (x, y)
∂x ∂y
1 1
√ p
= p 1 − x2 1 − y2
xy √ xy
1 − y2 − √ 1 − x2 − p
1 − x2 1 − y2
=0

Therefore u and v are functionally dependent.

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 50 / 66
 Example 5:
Verify whether the following functions are functionally dependent, and
if so, find the relation between them.
x +y
u= , v = tan−1 x + tan−1 y
1 − xy

Solution:
∂u ∂u
∂ (u, v )
= ∂∂ vx ∂∂ vy
∂ (x, y)
∂x ∂y
1 + y2 1 + x2
2 (1 − xy)2
= (1 − xy)
1 1
1 + x2 1 + y2
=0
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 51 / 66
 Example 5: (Contd.)

Hence u, v are functionally dependent.

Now

v = tan−1 x + tan−1 y
 
x +y
= tan−1
1 − xy
= tan−1 u
⇒ u = tan v .

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 52 / 66

 Example 5:

u = xy + yz + zx, v = x 2 + y 2 + z 2 and w = x + y + z, determine whether

there is a functional relationship between u, v , w and if so, find it.

Solution:
∂u ∂u ∂u
∂x ∂y ∂z
∂ (u, v , w) ∂v ∂v ∂v
=
∂ (x, y , z) ∂x ∂y ∂z
∂w ∂w ∂w
∂x ∂y ∂z
y +z z +x x +y y +z z +x x +y

= 2x 2y 2z =2 x y z

1 1 1 1 1 1

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 53 / 66

 Example 5: (Contd.)

x +y +z x +y +z x +y +z
∂ (u, v , w)
=2 x y z R1 → R1 + R2
∂ (x, y, z)
1 1 1
1 1 1

=2 x y z =0

1 1 1

Hence the functional relationship exists between u, v and w.

Now
w 2 = (x + y + z)2
= x 2 + y 2 + z 2 + 2(xy + yz + zx)
= v + 2u
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 54 / 66
 Example 6:

If u = y + z, v = x + 2z 2 and w = x − 4yz − 2y 2 , find the Jacobian of

u, v , w with respect to x, y, z. Comment on the result.

Solution:
∂u ∂u ∂u
∂x ∂y ∂z
∂ (u, v , w) ∂v ∂v ∂v
=
∂ (x, y, z) ∂x ∂y ∂z
∂w ∂w ∂w
∂x ∂y ∂z
0 1 1

= 1 0 4z

1 −4y − 4z −4y

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 55 / 66

 Example 6: (Contd.)

∂ (u, v , w)
= −1(−4y − 4z) + (−4y − 4z)
∂ (x, y , z)
=0

Therefore u, v and w are functionally dependent.

Now

v − w = 2z 2 + 4yz + 2y 2
= 2(y + z)2
= 2u 2 .

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 56 / 66

 Example 7:

∂ (u, v , w)
If u = xyz, v = xy + yz + zx and w = x + y + z, find .
∂ (x, y, z)

Solution:
∂u ∂u ∂u
∂x ∂y ∂z
∂ (u, v , w) ∂v ∂v ∂v
=
∂ (x, y , z) ∂x ∂y ∂z
∂w ∂w ∂w
∂x ∂y ∂z
yz xz xy

= y +z x +z y +z
1 1 1

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 57 / 66

 Example 7: (Contd.)

z(x − y) x(y − z) xy
∂ (u, v , w)
= x −y y −z y +x
∂ (x, y, z)
0 0 1
z x xy

= (x − y )(y − z) 1 1 y + x
0 0 1

= (x − y )(y − z)(z − x)

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 58 / 66

 Example 8:

∂ (u, v )
Find the value of the Jacobian , where u = x 2 − y 2 , v = 2xy and
∂ (r , θ )
x = r cos θ , y = r sin θ .

Solution:
∂ (u, v ) ∂ (u, v ) ∂ (x, y)
= ×
∂ (r , θ ) ∂ (x, y) ∂ (r , θ )
∂u ∂u ∂x ∂x
x y
= ∂ v ∂ v × ∂∂ yr ∂∂ θ
∂ ∂
y
∂x ∂y ∂r ∂θ
2x −2y cos θ −r sin θ
= ×
2y 2x sin θ r cos θ
= 4r 2 .r = 4r 3

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 59 / 66

 Example 9:

∂ (x, y, z)
If u = x + y + z, uv = y + z, uvw = z, show that = u2v .
∂ (u, v , w)

Solution: Given z = uvw

∂z ∂z ∂z
⇒ = vw, = vw, = uv
∂u ∂v ∂v
and y + z = uv ⇒ y + uvw = uv ⇒ y = uv − uvw
∂y ∂y ∂y
⇒ = v − vw, = u − uw, = −uv
∂u ∂v ∂v
Also u = x + y + z ⇒ x = u − y − z = u − uv − uvw ⇒ x = u − uv
∂x ∂x ∂x
⇒ = 1 − v, = −u, =0
∂u ∂v ∂v

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 60 / 66

 Example 9: (Contd.)

∂x ∂x ∂x
∂u ∂v ∂w
∂ (x, y , z) ∂y ∂y ∂y
=
∂ (u, v , w) ∂u ∂v ∂w
∂z ∂z ∂z
∂u ∂v ∂w
1−v −u 0
= v − vw u − uw −uv
vw uw uv
1 − vw −uw −uv
= v u 0 R1 → R1 + R2 ; R2 → R2 + R3
vw uw uv

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 61 / 66

 Example 9: (Contd.)

1 − vw −w −1
∂ (x, y, z)
= u2v v 1 0
∂ (u, v , w)
vw w 1
1 0 0
= u2v v 1 0 R1 → R1 + R3
vw w 1
= u2v

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 62 / 66

 Example 10:

∂ (x, y ) ∂ (r , θ )
If x = r cos θ , y = r sin θ verify that × = 1.
∂ (r , θ ) ∂ (x, y)

Proof:
y
Now x 2 + y 2 = r 2 and θ = tan−1
x
Differentiating partially w.r.t x, we get
∂r ∂θ 1  y 
2r = 2x and = . − 2
∂x ∂x y2 x
1+ 2
x
∂r x ∂θ y
⇒ = and =− 2
∂x r ∂x r
∂r y ∂θ x
Similarly ⇒ = and = 2
∂y r ∂y r

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 63 / 66

 Example 10: (Contd.)

∂r ∂r x y
2 2
∂x ∂y r r = x +y = 1
∂θ ∂θ = y x r3 r
− 2 2
∂x ∂y r r
and
∂x ∂x
∂r ∂ θ = cos θ −r sin θ
=r
∂y ∂y sin θ r cos θ
∂r ∂θ
∂x ∂x ∂r ∂r
∂ (x, y) ∂ (r , θ ) 1
× = ∂r ∂θ × ∂x ∂y
∂θ = r × r = 1
∂ (r , θ ) ∂ (x, y) ∂y ∂y ∂θ
∂r ∂θ ∂x ∂y
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 64 / 66
 Example 11:

∂ (x, y) ∂ (u, v )
If x = u(1 − v ), y = uv verify that × = 1.
∂ (u, v ) ∂ (x, y)

Proof:
Given x = u(1 − v ) ⇒ u = x + uv ⇒ u = x + y
∂u ∂u
⇒ = 1, =1
∂x ∂y
y y
and y = uv ⇒ v = ⇒ v =
u x +y
∂v y ∂v x
⇒ =− 2
, =
∂y (x + y) ∂ y (x + y )2
∂v y ∂v x
⇒ = − 2, = 2
∂y u ∂y u

Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 65 / 66

 Example 11: (Contd.)

∂u ∂u 1 1
∂ (u, v ) x +y 1
= ∂∂ vx ∂y =
∂v y x = u2 = u
∂ (x, y) − 2
∂x ∂y u u2
and
∂x ∂x
∂ (x, y ) ∂ v = 1 − u −u = u
= ∂u
∂ (u, v ) ∂y ∂y v u
∂u ∂v
∂x ∂x ∂u ∂u
∂ (x, y) ∂ (u, v ) 1
× = u
∂ ∂v × ∂x
∂v
∂y
∂v = u × u = 1
∂ (u, v ) ∂ (x, y ) ∂y ∂y
∂u ∂v ∂x ∂y
Dr. S. SABARINATHAN, Dept. of Mathematics, KTR () Unit II-Maxima and Minima 66 / 66