Dr.kamlendra Kumar 1
Subject Name:- Engineering Mathematics
Subject Code:- KAS 103T
Unit No.:-3
Lecture No.:- 7
Topic Name :- Differential Calculus-II
Topic Name: Maxima and Minima contd..
Dr. Kamlendra Kumar
Department of Mathematics

Contents
• Extrema of function of two variables
• Lagrange’s Method of Multipliers
• Problems
2
Dr.kamlendra Kumar

Solved Problems
Dr.kamlendra Kumar 3
Example . Find the minimum distance from the point (1,2,0) to
2 2 2
the cone z .
Sol. Let x,y,z be any point on cone . The distance D from (1,2,0)
is
2 2 2
( 1) ( 2) ( 0) ,
2 2
1 2
x y
D x y z
f(x,y,z) (x ) (y )
 
     
    2...(1)
2 2 2
(x,y,z) x 0 ...(2)
Let the auxilliary equation be
2 2 2 2 2 2
F(x,y,z) [ 1 2 ] (x ) 0 ...(3)
(x-1)
2( 1) 2 0 - ...(4)
x
z
y z
(x ) (y ) z y z
F
x x
x


 

   
        

     


Solved Problems
Dr.kamlendra Kumar 4
(y-2)
2( 2) 2 0 - ...(5)
y
2 2 0 1 ...(6)
(x-1) (y-2)
From equations (4), (5), (6), - 1
x y
1
we get x , y 1
2
Substituting in (2) , we ge
F
y y
y
F
z z
z
 
 

     


    

  
  
5
t
2
5
Minimum distance from (1,2,0)
2
z  


Solved Problems
Dr.kamlendra Kumar 5
2 2 2
Example. Find the minimum value of z , given that
ax by cz p.
2 2 2
Sol. Let z ...(1)
(x,y,z) ax by cz p ...(2)
Let the auxilliary equation be,
2 2 2
F(x,y,z) z (ax by cz p) 0 ...(3
x y
f(x,y,z) x y
x y


 
  
  
   
        )
2x
2 0 - ...(4)
a
(2y)
2 0 - ...(5)
b
F
x a
x
F
y b
y
 
 

    


    


Solved Problems
Dr.kamlendra Kumar 6
2z
2 0 ...(6)
c
2x 2y 2z
From equations (4), (5), (6), -
a b c
we get z , y
Substituting in (2) , we get
ap bp cp
, ,
2 2 2 2 2 2 2 2 2
a a a
Minimum v
F
z c
z
cx bx
a a
x y z
b c b c b c
 

     

   
  
  
     
2
p
alue of function is
2 2 2
a b c

 

Solved Problems
Dr.kamlendra Kumar 7
Example. Find the dimensions of a rectangular box of maximum
capacity whose surface area is given when (i) box is open at the top
(ii) box is closed.
Sol. Let , , be the dimensions of the re
f(x,y,z) x y z
 ctangular box
Volume V xyz ....(1)
Total surface area of the box is when it is open
S xy 2yz 2zx a ...(2)
Let the auxilliary equation be,
F(x,y,z) (xy 2yz 2zx a) ...(3)
xyz
F
yz
x


   
    



yz
( 2 ) 0 - ...(4)
(y 2z)
y z
 
    


Solved Problems
Dr.kamlendra Kumar 8
xz
( 2 ) 0 - ...(5)
(x 2z)
xy
(2 2 ) 0 ....(6)
(2y 2x)
yz xz xy
From equations (4), (5), (6), -
(y 2z) (x 2z) (2y 2x)
we get z , y
2
Subst
F
xz x z
y
F
xy y x
z
x
x
 
 

     
 

      
 
   
  
  
a a 1 a
2
ituting in (2) , we get , x y ,
3 3 2 3
a 1 a
The dimensions are x y , z
3 2 3
a
When box is closed , x y z
3
x z
    
  
  

Solved Problems
Dr.kamlendra Kumar 9
Example. Show that if the perimeter of a triangle is constant,
its area is a maximum when it is equilateral.
Sol. Let be the sides of the triangle,then its perimeter is
2 c k .
a,b,c
s a b
    ..(1)
2
Area of triangle A s(s-a)(s-b)(s-c) ...(2)
Lagrange's function is F s(s-a)(s-b)(s-c) ( c k)
For stationary values dF 0
s(s-b)(s-c) 0 s(s-b)(s-c) ...(4)
a b
F
a
F
b

 

    


     



s(s-a)(s-c) 0 s(s-a)(s-c) ...(5)
 
     

Solved Problems
Dr.kamlendra Kumar 10
s(s-a)(s-b) 0 s(s-b)(s-a)...(6)
From (4), (5), (6) we get a b c., hence the triangle is equilateral.
2s
From (1), we get a
3
c 2s-a-b
2
A ( )( )( )
2
now, 2 ( )
2
2
now,
F
z
b c
s s a s b a b s f
f
s s b
a
 

     

 
  

     

  

 2
2 ( ) and (2 2 3 )
2
f f
s s a s a b s
a b
b

     
 


Solved Problems
Dr.kamlendra Kumar 11
2 2 2 2
2s s
at stationary point , - 0,
2 3 3
2 2
2s
and -
2 3
2
2 2 2 4 2
s
. 0 and 0
2 2 2
3
Hence, area is maximum when a b c.
Ex.5. In a plane triangle ABC,find
f f
a b
a
f
b
f f f f
a b
a b a
 
 
 
   
 
 
  



 
   
 
   
 
 
  
 
 
the maximum value of
cosAcosBcosC.
Sol. f(A,B,C) cos cos cos ...(1)
(A,B,C) A B C- 0...(2)
A B C
 

   

Solved Problems
Dr.kamlendra Kumar 12
Lagrange's function F(A,B,C) cos cos cos
( )...(3)
For stationary values dF 0
F F
sin cos cos 0, sin cos cos 0,
A B
F
sin cos cos 0
C
we get tanA tanB tanC
o A B C
f
A B C
A B C
A B C B A C
C A V
r
 

   

 
     
 

  

 
 
rom (2) A B C
3
1
Hence maximum value of function cos cos cos .
3 3 3 8

  
  
 

Important Questions
Dr.kamlendra Kumar 13
Q.1. Find the volume of the largest rectangular parallelopiped that
2 2 2
x y z
can be inscribed in the ellipsoid + + =1.
2 2 2
a b c
2
Q.2. Find the largest product of the numbers x,y,z when x+y+z =16.
Q.3. A rectangular box, open at the top, is to have a volume of 32 cc..
Find the dimensions of the box which requires least amount of material
for its construction.
4 2 2 4
Q.4. Examine for extreme values: - - -1 .
Q
x x y y

2
.5. The temperature T at any point (x,y,z) in space is T(x,y,z) =kxyz
where k is a constant.Find the highest temperature on the surface of the
2 2 2 2
sphere x +y +z =a .

References
Reference Books:-
(i) B.V.Ramanna , Higher Engineering Mathematics (T.M.H. Publ.)
(ii) R.K. Jain & R.K. Iyenger, Advance Engineering Mathematics,
Narosa Publishing House.2002.
Text Books:-
(i) E. Kreyszig, Advance Engineering Mathematics,John Wiley &
Sons
(ii) H.S.Gangwar and Prabhakar Gupta (New Age International
Publishers)
(iii) B.S. Grewal, Higher Engineering Mathematics, Khanna
Publisher,2005
Dr.kamlendra Kumar 14

Thank You
15
Dr.kamlendra Kumar