Lagranges method of undetermined multiplers examples

1. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
Lagrange’s Method of Undetermined Multipliers
Suppose we require to find the maximum and minimum values of 𝑓(𝑥, 𝑦) where
𝑥, 𝑦, 𝑧 are subject to a constraint equation
𝑔(𝑥, 𝑦, 𝑧) = 0
We define a function
𝐹(𝑥, 𝑦, 𝑧, 𝜆 ) = 𝑓(𝑥, 𝑦, 𝑧) + 𝜆 𝑔(𝑥, 𝑦, 𝑧) … (1)
Where 𝜆 is called Lagrange Multiplier which is independent of 𝑥, 𝑦, and 𝑧.
The necessary conditions for a maximum or minimum are
𝜕𝐹
𝜕𝑥
= 0 … .(2)
𝜕𝐹
𝜕𝑦
= 0 … . (3)
𝜕𝐹
𝜕𝑧
= 0 … . (4)
Solving the four equations for four unknowns λ ,𝑥, 𝑦, 𝑧, we obtain the point (𝑥, 𝑦, 𝑧). The
point may be a maxima, minima or neither which is decided by the physical consideration.
This method is also applicable when we have more than one constraint equation
connecting the variables.
Example:
𝟏 𝟏
𝒙 𝒚 𝒛
Find the minimum value of 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 subject to the condition
𝟏
+ + = 𝟏
Solution:
Let the auxiliary function ‘F’ be
𝑥 𝑦 𝑧
F(x,y,z,λ ) =(𝑥2 + 𝑦2 + 𝑧2) + 𝜆 (1
+ 1
+ 1
− 1)
Where 𝜆 is Lagrange Multiplier
𝜕𝐹 −1
𝜕𝑥
= 2𝑥 + 𝜆 (
𝑥2 )
= 2𝑥 − 𝜆
𝑥2
𝜕𝐹 −1
𝜕𝑦
= 2𝑦 + 𝜆 (
𝑦2 )
= 2𝑦 − 𝜆
𝑦2
𝜕𝐹 −1
𝜕𝑧
= 2𝑧 + 𝜆 (
𝑧2 )
= 2𝑧 − 𝜆
𝑧2
For a minimum at (𝑥, 𝑦, 𝑧) we have
𝜕𝐹
𝜕𝑥
= 0
𝜕𝐹
𝜕𝑦
= 0
𝜕𝐹
𝜕𝑧
= 0
𝜆
2𝑥 −
𝑥2 = 0
𝜆
2𝑥 =
𝑥2
𝜆
2𝑦 −
𝑦2 = 0
𝜆
2𝑦 =
𝑦2
𝜆
2𝑧 −
𝑧2 = 0
𝜆
2𝑧 =
𝑧2
MA8151 ENGINEERING MATHEMATICS I

2. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
𝜆
𝑥3 =
2
1
𝑥 =
𝜆
)3
….(1)
(
2
𝜆
𝑦3 =
2
1
𝑦 = (
𝜆
)3
….(2)
2
𝜆
𝑧3 =
2
1
𝑧 =
𝜆
)3
….(3)
(
2
From (1), (2) and (3), we get
𝑥 = 𝑦 = 𝑧
1 1 1
𝑥 𝑦 𝑧
Given: + + = 1
∴ 3 1
= 1
𝑥
∴ 3 = 𝑥
∴ ⇒ 𝑦 = 3 and 𝑧 = 3
∴ (3, 3, 3) is the point where minimum values occur.
The minimum value is 32 + 32 + 32 = 9+ 9 + 9 =27.
Example:
A rectangular box open at the top, is to have a volume of 32cc. find the dimensions
of the box that requires the least material for its construction.
Solution:
Let 𝑥, 𝑦, 𝑧 be the length, breadth and height of the box.
Surface area = 𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥
Volume = 𝑥𝑦𝑧 = 32
Let the auxiliary function F be
F(𝑥, 𝑦, 𝑧, λ ) =(𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥 ) + 𝜆(𝑥𝑦𝑧 − 32)
Where 𝜆 is Lagrange Multiplier
𝜕𝐹
𝜕𝑥
= 𝑦 + 2𝑧 + 𝜆𝑦𝑧
𝛛𝐹
= 𝑥 + 2𝑧 + 𝜆𝑥𝑧
𝛛𝑦
𝛛𝐹
= 2𝑥 + 2𝑦 + 𝜆𝑥𝑦
𝛛𝑧
When F is extremum
𝜕𝐹
𝜕𝑥
= 0
𝜕𝐹
𝜕𝑦
= 0
𝜕𝐹
𝜕𝑧
= 0
𝑦 + 2𝑧 + 𝜆𝑦𝑧 = 0
⇒ y + 2 z = − 𝜆𝑦𝑧
𝑥 + 2𝑧 + 𝜆𝑧𝑥 = 0
⇒ x + 2 z = − 𝜆𝑦𝑧
2𝑥 + 2𝑦 + 𝜆𝑥𝑦 = 0
⇒ 2x + 2 y = − 𝜆𝑥𝑦
MA8151 ENGINEERING MATHEMATICS I

3. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
⇒ 1
+ 2
= − 𝜆 …(1)
𝑧 𝑦
⇒ 1
+ 2
= − 𝜆 …(2)
𝑧 𝑥
⇒ 2
+ 2
= − 𝜆 …(3)
𝑦 𝑥
From (1) and (2), we get
1 2 1 2
𝑧
+
𝑦
=
𝑧
+
𝑥
2 2
𝑦
=
𝑥
𝑥 = 𝑦 … (4)
From (2) and (3), we get
1 2 2 2
𝑧
+
𝑥
=
𝑦
+
𝑥
1 2
𝑧
=
𝑦
𝑦 = 2𝑧 … (5)
From (4) and (5), we get
𝑥 = 𝑦 = 2𝑧
Volume = 𝑥𝑦𝑧 = 32 ⇒ (2z) (2z) z = 32 ⇒ 4 𝑧3 = 32
𝑧3 =
32
4
= 8 ⇒ 𝑧 = 2 𝑖. 𝑒. 𝑥 = 4, 𝑦 = 4, 𝑧 = 2
∴ Cost minimum when 𝑥 = 4, 𝑦 = 4, 𝑧 = 2
Example:
A rectangular box open at the top is to have a given capacity K. Find the
dimensions of the box requiring least material for its construction.
Solution:
Let 𝑥, 𝑦, 𝑧 be the dimensions of the box.
Surface area = 𝑥 𝑦 + 2𝑦𝑧 + 2𝑧𝑥
Volume = 𝑥𝑦𝑧 = 𝐾
Let the auxiliary function F be
F(x,y, z, 𝜆 ) =(𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥 ) + 𝜆(𝑥𝑦𝑧 − 𝑘)
Where 𝜆 is Lagrange Multiplier
𝜕𝐹
𝜕𝑥
= 𝑦 + 2𝑧 + 𝜆𝑦𝑧
𝛛𝐹
= 𝑥 + 2𝑧 + 𝜆𝑧𝑥
𝛛𝑦
𝛛𝐹
= 2𝑥 + 2𝑦 + 𝜆𝑥𝑦
𝛛𝑧
When F is extremum.
𝜕𝐹
𝜕𝑥
= 0
𝜕𝐹
𝜕𝑦
= 0
𝜕𝐹
𝜕𝑧
= 0
MA8151 ENGINEERING MATHEMATICS I

4. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
𝑦 + 2𝑧 + 𝜆𝑦𝑧 = 0
⇒y + 2 z = − 𝜆𝑦𝑧
⇒ 1
+ 2
= −…(1)
𝑧 𝑦
𝑥 + 2𝑧 + 𝜆𝑧𝑥 = 0
⇒ x + 2 z = − 𝜆𝑧𝑥
⇒ 1
+ 2
= − 𝜆 ...(2)
𝑧 𝑥
2𝑥 + 2𝑦 + 𝜆𝑥𝑦 = 0
⇒2x + 2 y = − 𝜆𝑥𝑦
⇒ 2
+ 2
= − 𝜆 ...(3)
𝑦 𝑥
From (1) and (2), we get
1 2 1 2
𝑧
+
𝑦
=
𝑧
+
𝑥
2 2
𝑦
=
𝑥
𝑥 = 𝑦 … . . (5)
From (2) and (3), we get
1 2 2 2
𝑧
+
𝑥
=
𝑦
+
𝑥
1 2
𝑧
=
𝑦
𝑦 = 2𝑧 … . (6)
From (4) and (5), we get
𝑥 = 𝑦 = 2𝑧
∴ Volume = 𝑥𝑦𝑧 = k ⇒ (2z) (2z) z = k ⇒ 4 𝑧3 = 𝑘
4 𝑧3 = 𝑘 ⇒ 𝑧3 = 𝑘
4
1 1 1
𝑘 3 𝑘 3 𝑘 3
𝑧 = (
4
) ; 𝑥 = 2 (
4
) ; 𝑦 = 2 (
4
)
∴ Value of minimum =
4
𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥
1
= 4 (𝑘
)3
4
1
+ 4 (𝑘
)3
+ 4
4
1
(𝑘
)3
2
= 12 (𝑘
)3
4
= 3 (2 𝑘 )2⁄3
MA8151 ENGINEERING MATHEMATICS I
Example:
Find the point on the plane a x + b y + c z =p at which 𝒇 = 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 has a
stationary value and find the stationary value of f , using Lagrange’s method of
multipliers
Solution:
Let the auxiliary function F be
F(x,y,z,λ ) = 𝑥2 + 𝑦2 + 𝑧2 + 𝜆(𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 − 𝑝)
Where 𝜆 is Lagrange Multiplier

5. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
𝜕𝐹
𝜕𝑥
= 2𝑥 + 𝜆𝑎
𝜕𝐹
𝜕𝑦
= 2𝑦 + 𝜆 𝑏
𝜕𝐹
𝜕𝑧
= 2𝑧 + 𝜆 𝑐
When F is extremum.
𝜕𝐹
𝜕𝑥
= 0
𝜕𝐹
𝜕𝑦
= 0
𝜕𝐹
𝜕𝑧
= 0
2𝑥 + 𝜆 𝑎 = 0
⇒ 2 𝑥 = − 𝜆 𝑎
⇒ 𝑥
= − 𝜆
… (1)
𝑎 2
2𝑦 + 𝜆 𝑏 = 0
⇒2 𝑦 = − 𝜆 𝑏
⇒ 𝑦
= − 𝜆
… (2)
𝑏 2
2𝑧 + 𝜆 𝑎 = 0
⇒2𝑧 = − 𝜆 𝑎
⇒ 𝑧
= − 𝜆
…. (3)
𝑐 2
From (1), (2) & (3), we get
=
𝑏2 =
𝑥 𝑦 𝑧
𝑎
=
𝑏
=
𝑐
𝑎 𝑥 𝑏 𝑦 𝑐 𝑧
𝑎2 𝑐2
⇒ = ⇒
𝑎 𝑥 + 𝑏𝑦 + 𝑐𝑧
=
𝑝
𝑎2 +
𝑎 𝑥 𝑏 𝑦 𝑐 𝑧
𝑎2 =
𝑏2 𝑐2
𝑎𝑝
𝑥 = ; 𝑦 =
𝑎2 + 𝑏2 + 𝑐2
𝑏𝑝
𝑎2 + 𝑏2 + 𝑐2 𝑎2 + 𝑏2 + 𝑐2
; 𝑧 =
𝑏2 + 𝑐2
𝑐𝑝
𝑎2 + 𝑏2 + 𝑐2
Stationary value of 𝑓 = 𝑥2 + 𝑦2 + 𝑧2
= (
2 2
𝑎 𝑝 𝑏 𝑝 𝑐 𝑝
𝑎2 + 𝑏2 + 𝑐2 ) + ( 𝑎2 + 𝑏2 + 𝑐2 ) + ( 𝑎2 + 𝑏2 + 𝑐2
)
2
=
𝑎2 𝑝2 + 𝑏 2 𝑝2+𝑐2 𝑝2
( 𝑎2 + 𝑏2+ 𝑐2)2
(𝑎2+𝑏2+𝑐2)𝑝2
= =
𝑝2
(𝑎2+𝑏2+𝑐2)2 𝑎2 + 𝑏2 + 𝑐2
Example:
Find the dimensions of the rectangular box without top of maximum capacity with
surface area 432 square metre
Solution:
Let 𝑥, 𝑦, 𝑧 be the length, breadth and height of the box.
Surface area = 𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥 = 432
Volume = 𝑥𝑦𝑧
Let the auxiliary function F be
MA8151 ENGINEERING MATHEMATICS I

6. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
F( , 𝑦, 𝑧 ,λ )= 𝑥𝑦𝑧 + 𝜆 (𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥 − 432 )
𝜕𝐹
𝜕𝑥
= 𝑦𝑧 + 𝜆 (𝑦 + 2𝑧)
𝜕𝐹
𝜕𝑦
= 𝑥𝑧 + 𝜆 (𝑥 + 2𝑧)
𝜕𝐹
𝜕𝑧
= 𝑥𝑦 + 𝜆 (2𝑦 + 2𝑥)
To find the stationary value.
𝐹𝑥 = 0 𝐹𝑦 = 0 𝐹𝑧 = 0
𝑦𝑧 + 𝜆 (𝑦 + 2𝑧) = 0
⇒ 𝑦𝑧 = −𝜆 (𝑦 + 2𝑧)
𝑦 + 2𝑧 −1
⇒
𝑦𝑧
=
𝜆
1 2 1
⇒
𝑧
+
𝑦
= −
𝜆
… (1)
𝑥𝑧 + 𝜆 (𝑥 + 2𝑧) = 0
⇒ 𝑥𝑧 = −𝜆 (𝑥 + 2𝑧)
𝑥 + 2𝑧 −1
⇒
𝑥𝑧
=
𝜆
1 2 1
⇒
𝑧
+
𝑥
= −
𝜆
… (2)
𝑥𝑦 + 𝜆 (2𝑦 + 2𝑥) = 0
⇒ 𝑥𝑦 = −𝜆 (2𝑦 + 2𝑥)
2𝑦 + 2𝑥 −1
⇒
𝑥𝑦
=
𝜆
2 2 1
⇒
𝑥
+
𝑦
= −
𝜆
… (3)
𝐹𝑟𝑜𝑚 (1) & (2),𝑤𝑒 𝑔𝑒𝑡
1 2 1 2
𝑧
+
𝑦
=
𝑧
+
𝑥
2 2
⇒
𝑦
=
𝑥
⇒ 𝑥 = 𝑦 …( 4 )
𝐹𝑟𝑜𝑚 (2) & (3),𝑤𝑒 𝑔𝑒𝑡
1 2 2 2
𝑧
+
𝑥
=
𝑦
+
𝑥
1 2
⇒
𝑧
=
𝑦
⇒ 𝑦 = 2𝑧 …( 5 )
From (4) & (5), we get 𝑥 = 𝑦 = 2𝑧
Surface area = 𝑥𝑦 + 2𝑦𝑧 + 2𝑧𝑥 = 432
(2𝑧)(2𝑧) + 2 (2𝑧)𝑧 + 2𝑧 (2𝑧) = 432
4𝑧2 + 4𝑧2 + 4𝑧2 = 432
12𝑧2 = 432
𝑧2 = 36 ∴ 𝑧 = 6
∴ 𝑥 = 12 , 𝑦 = 12, 𝑧 = 6 𝑏𝑦 (6)
Thus, the dimension of the box is 12, 12, 6.
Maximum volume = 12 x 12 x 6 = 864 cubic metres.
MA8151 ENGINEERING MATHEMATICS I

7. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
Example:
The temperature 𝒖( 𝒙,𝒚, 𝒛 ) at any point in space is 𝒖 = 𝟒𝟎𝟎𝒙𝒚𝒛𝟐 .Find the highest
temperature on surface of the sphere 𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 = 𝟏 .
Solution:
𝑢 = 𝑓 = 400𝑥𝑦𝑧2 … (𝐴)
∅ = 𝑥2 + 𝑦2 + 𝑧2 − 1 = 0
Let the auxiliary function F be
F( 𝑥, 𝑦, 𝑧 , 𝜆 ) =
… (𝐵)
400𝑥𝑦𝑧2 + 𝜆 (𝑥2 + 𝑦2 + 𝑧2 − 1)
To
find the stationary value.
𝐹𝑥 = 0 𝐹𝑦 = 0 𝐹𝑧 = 0
400𝑦𝑧2 + 𝜆 (2𝑥) = 0
400𝑦𝑧2 = −𝜆 (2𝑥)
200𝑦𝑧2
𝑥
= −𝜆 … (1)
400𝑥𝑧2 + 𝜆 (2𝑦) = 0
400𝑥𝑧2 = −𝜆 (2𝑦)
200𝑥𝑧2
𝑦
= −𝜆 … (2)
800𝑥𝑦𝑧 + 𝜆(2𝑧) = 0
800𝑥𝑦𝑧 = − 𝜆(2𝑧)
400𝑥𝑦 = − 𝜆 … (3)
From (1) & (2) , we get 𝑦2 = 𝑥2 … . (4)
From (2) & (3) , we get 𝑧2 = 2𝑦2 …. (5)
From (4) & (5), we get
2
𝑥2 = 𝑦2 = 1
𝑧2 …(6)
2
(𝐵) ⇒ 1
𝑧2 + 1
𝑧2 + 𝑧2 − 1 = 0
2
⇒ 2𝑧2 = 1
⇒ 𝑧2 = 1
⇒ 𝑧 = ± 1
2 2
(6) ⇒ 𝑥2 = 1
(1
) = 1
⇒ 𝑥 = ± 1
2 2 4 2
(6) ⇒ 𝑦2 = 1
(1
) = 1
⇒ 𝑦 = ± 1
2 2 4 2
∴ 𝑢 = 400𝑥𝑦𝑧2 , we select 𝑥, 𝑦, 𝑧 to be positive
1 1 1
⇒ 𝑢 = 400 (
2
) (
2
) (
2
)
⇒ 𝑢 = 50
𝜕𝐹
= 400𝑦𝑧2 + 𝜆 (2𝑥)
𝜕𝑥
𝜕𝐹
= 400𝑥𝑧2 + 𝜆 (2𝑦)
𝜕𝑦
𝜕𝐹
= 800𝑥𝑦𝑧 + 𝜆(2𝑧)
𝜕𝑧
MA8151 ENGINEERING MATHEMATICS I

8. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
∴ Maximum temperature is 50
Example:
Find the maximum volume of the largest rectangular parallelepiped that can be
𝒂𝟐 𝒃𝟐 𝒄𝟐
𝟐 𝟐 𝟐
inscribed in an ellipsoid 𝒙
+ 𝒚
+ 𝒛
= 𝟏
Solution:
Let a vertex of such parallelepiped be (x, y, z)
Then all the vertices will be ( ±𝑥 , ±𝑦, ±𝑧 )
Then, the sides of the solid be 2𝑥, 2𝑦, 2𝑧 (lengths)
Hence, the volume 𝑉 = (2𝑥) (2𝑦) (2𝑧) = 8𝑥𝑦𝑧
Let 𝑓 = 8𝑥𝑦𝑧
2 2 2
∅ = 𝑥
+ 𝑦
+ 𝑧
− 1 = 0
𝑎2 𝑏2 𝑐2
Let the auxiliary function F be
𝑎2 𝑏2 𝑐2
2 2 2
F( 𝑥, 𝑦, 𝑧 , 𝜆 ) = 8𝑥𝑦𝑧 + 𝜆 (𝑥
+ 𝑦
+ 𝑧
− 1)
𝜕𝐹 2𝑥
𝜕𝑥
= 8𝑦𝑧 + 𝜆
𝑎2
𝜕𝐹 2𝑦
𝜕𝑦
= 8𝑥𝑧 + 𝜆
𝑏2
𝜕𝐹 2𝑧
𝜕𝑧
= 8𝑥𝑦 + 𝜆
𝑐2
To find the stationary value.
𝐹𝑥 = 0 𝐹𝑦 = 0 𝐹𝑧 = 0
2𝑥
8𝑦𝑧 + 𝜆
𝑎2 = 0
2𝑥
⇒ 8𝑦𝑧 = −𝜆
𝑎2
Xly 𝑥
⇒ 4𝑥𝑦𝑧
= 𝑥2
…(1)
2 −𝜆 𝑎2
2𝑦
8𝑥𝑧 + 𝜆
𝑏2 = 0
2𝑦
⇒ 8𝑥𝑧 = −𝜆
𝑏2
Xly 𝑦
⇒ 4𝑥𝑦𝑧
= 𝑦2
…(2)
2 −𝜆 𝑏2
2𝑧
8𝑥𝑦 + 𝜆
𝑐2 = 0
2𝑧
⇒ 8𝑥𝑦 = −𝜆
𝑐2
Xly 𝑧
⇒ 4xyz
= z2
…(3)
2 −λ c2
From (1), (2) & (3), we get
a2 b2 c2
x2
= y2
= z2
… (4)
a2 b2 c2
2 2 2
Given x
+ y
+ z
= 1
a2
2
⇒ 3x
= 1 by (4)
2
⇒ x2 = a
⇒ x = a
3 √3
Similarly, = b
; 𝑧 = c
√3 √3
MA8151 ENGINEERING MATHEMATICS I

9. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
The extremum point is ( a
,
√3 √3 √3
b
, c
)
3√3
Maximum volume V= 8(a bc
)
Example:
Find the minimum values of 𝐱𝟐𝐲𝐳𝟑 subject to the condition
Solution:
Let 𝑓 = 𝑥2𝑦𝑧3
∅ = 2𝑥 + 𝑦 + 3𝑧 − 𝑎 = 0
Let the auxiliary function F be
𝟐𝒙 + 𝒚 + 𝟑𝒛 = 𝒂.
F( 𝑥, 𝑦, 𝑧 , 𝜆 ) = 𝑥2𝑦𝑧3 + λ (2x + y + 3z − a)
𝜕F
= 2𝑥yz3 + λ2
𝜕x
𝜕F
= x2z3 + λ
𝜕y
𝜕F
= 3 x2yz2 + 3λ
𝜕z
To find the stationary value.
Fx = 0 Fy = 0 Fz = 0
2xyz3 + λ2 = 0
xyz3 = − λ … (1)
x2z3 + λ = 0
x2z3 = − λ … (2)
3 x2yz2 + 3λ = 0
x2yz2 = −λ … (3)
From (1) & (2),we get
xyz3 = x2z3
𝑥 = 𝑦 … (4)
From (2) & (3),we get
x2z3 = x2yz2
𝑧 = 𝑦 … (5)
From (4) & (5), we get
𝑥 = 𝑦 = 𝑧 … (6)
Given: 2x + y + 3z = a
⇒ 2z + z + 3z = a
⇒ 6𝑧 = 𝑎
a
6
⇒ z = (6) ⇒ x = y = z =
a
6
∴ The stationary point is ( ,
a a a
6 6 6
, )
MA8151 ENGINEERING MATHEMATICS I
Hence, Minimum value of f = x2yz3

10. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
a a a
= ( ) ( ) ( )
2 3 a
6 6 6 6
= ( )
6
Example:
Find the maximum value of𝐱𝐦𝐲𝐧𝐳𝐩, when 𝐱 + 𝐲 + 𝐳 = 𝐚.
Solution:
Let 𝑓 = 𝑥𝑚𝑦𝑛𝑧𝑝
∅ = 𝑥 + 𝑦 + 𝑧 − 𝑎 = 0
Let the auxiliary function F be
F( x, y, z ,λ ) = xmynzp + λ (x + y + z − a)
𝜕F
= mxm−1ynzp + λ
𝜕x
𝜕F
= nxmyn−1zp + λ
𝜕y
𝜕F
= pxmynzp−1 + λ
𝜕z
To find the stationary value.
Fx = 0 Fy = 0 Fz = 0
mxm−1ynzp + λ = 0
⇒ 𝑚xm−1ynzp = − λ
mxmynzp
x
= − λ… (1)
nxmyn−1zp + λ = 0
⇒ 𝑛xmyn−1zp = − λ
nxmynzp
y
= − λ … (2)
pxmynzp−1 + λ = 0
⇒ 𝑝xmynzp−1 = − λ
pxmynzp
z
= − λ …(3)
From (1), (2) & (3), we get
mxmy𝚗zp
= nxmy𝚗zp
= pxmy𝚗zp
x y z
x y z
÷ xmynzp ⇒ m
= n
= p
… (4)
x = m
z … (5) y = n
z … (6)
p p
Given: 𝑥 + 𝑦 + 𝑧 = 𝑎
p
m n
p
⇒ z + z + z = a (5) ⇒ x =
m ap
p m+n+p
am
m n
p p
⇒ [ + + 1] z = a =
⇒ [
m+n+p
p
]z = a (5) ⇒ y =
m+n+p
n ap
p m+n+p
⇒ z =
ap
m+n+p
=
an
m+n+p
MA8151 ENGINEERING MATHEMATICS I

11. ROHINI COLLEGE OF ENGINEERING AND TECHNOLOGY
∴ The stationary point is (
am an
, ,
m+n+p m+n+p m+n+p
ap
)
m+n+p m+n+p
m n
∴ The Maximum value of f = ( am
) ( an
) (
ap
m+n+p
)
p
m+𝚗+p m 𝚗 p
Max. Value of f = a m n p
(m+n+p)m+𝚗+p
MA8151 ENGINEERING MATHEMATICS I