### KAS 103_Unit 3_Lecture _7.pptx

1. 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
2. Contents • Extrema of function of two variables • Lagrange’s Method of Multipliers • Problems 2 Dr.kamlendra Kumar
3. 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                          
4. 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   
5. 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                  
6. 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   
7. 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        
8. 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           
9. 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)        
10. 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          
11. 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       
12. 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         
13. 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 .
14. 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
15. Thank You 15 Dr.kamlendra Kumar