Maxima and Minima for Functions with side conditions. Lagrange s Multiplier. Question Find the critical points of w= xyz subject to the condition

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1 Maima and Minima for Functions with side conditions. Lagrange s Multiplier. Find the critical points of w= z subject to the condition + + z =. We form the function ϕ = f + λg = z+ λ( + + z ) and obtain four equations ϕ = z+ λ= ϕ = z+ λ = ϕ = + λ z= z & + + z = Multipling the first three equations b z,, respectivel, adding z and using in fourth equation we find λ =. Using this relation we have (,, ± ), (, ±,), ( ±,,), and ±, ±, ± as the critical points. Find the critical points of the function where + = 5. Test for maima & minima. F(,, λ) = λ( + 5) = + + λ=. (i) F = + 6+ λ = (ii) z = & + 5=... (iii) (i) ( + λ) + =... (iv) (ii) + (8 + λ) =.... (v) Multipling equation (iv) b, (v) b ( + λ) and adding ( + λ) + = ( + λ) + ( + λ)(8 + λ) = ( + λ)(8 + λ) = = or λ + 9λ 6= =, λ = 8, 7 From (ii), = = (,) does not satisf (iii) It is not a critical point. λ = 8 = form (iv) Put this value of in (iii)

2 6 + = 5 9 =± (,) & (, ) are the critical points. Similarl when λ = 7, we have = from (iv) And putting the value of in (iii) we get =± ( ±, ± ) are the other two critical points. A= = + λ B= F = C= F = 6+ λ When λ = 8 A = + 6= 8, B =, C = 6+ 6= and so B AC = = F(,, λ ) = ( + 5) when λ = 8 F = + + At (,) F = F( + h, + h) F(,) (, ) 9 6 = ( h) ( h)( h) 6( h) 9( ) + ( )( ) 6() + = 9(6 8 h+ h ) + ( h h ) + 6(9+ 6 h+ h ) + 88 = 7h+ 9h + h h h+ 6h + 88 = 9h (,) is the point of minimum value. Similarl (, ) gives a point of minimum value. And when λ = 7, ( ±, ± ) are the point of maimum value. Find the critical points of w= + z, where Test for a maima and minima. Consider the function F( z,, ) = + z+ λ( + + z ) = + λ, F = λ, F = + λz z + + z =. For critical points, we have + λ =... (i) λ =... (ii) + λz =.. (iii) and + + z = (iv) Solving these equations we have λ =± λ = gives,, as the critical point. λ = gives,, as the critical point.

3 A= = λ, B= F =, C= F = λ i) λ = A =, B =, C = so B AC = < and A >,, is a point of relative minimum value. ii) λ = A =, B =, C = so B AC = < and A <,, is a point of relative maimum value. Find the critical points of w= z where z =. Test for the maima and minima. Consider F = z+ λ( + ) + λ( z) For critical points = z+ λ+ λ =. (i) F = z+ λ =... (ii) z F = λ =.. (iii) & + =... (iv) z =.. (v) z From (iii) λ = and from (ii) λ = Putting in (i), we get z z + = z z+ = = z form (iv) But from (iv), = = = The critical points are (,, ) and (,,). From (ii) i) At A F λ = =, B AC = z λ + = = ( ) = ( ) ±,, ±, B= F = z, C= F = λ + = & =, ± ±,, ±, z z z λ = B AC = z = ±,, ±, we have B ( ) ( ) AC = <

4 z And A= F = λ = = < Function is maimum at ±,, ±. Similarl we can show that w is maimum at (,,) and minimum at ±,, ± & (,,). Find the point to the curves + z =, + = nearest to the origin (,,,). Let ( z,, ) be a point on the curve. Then its distance from the origin is given b We are to minimize + + z f = d = + + z subject to the conditions + z =, + = Consider F = + + z + λ + z + λ + = + ( ) λ+ λ F = + ( ) λ + λ ( ) ( ) Fz = z+ λ( ) z For critical points, we have ( + λ+ λ) λ= (i) ( + λ+ λ) λ=.(ii) ( z λ ) =... (iii) + z =. (iv) + =.... (v) From (iii), we have z = & λ =. z = in (iv) gives + = = + But + = = = or = or both are zero. We can not take =, = at a same time because it gives (,,) which is origin itself. z =, = in (v) & z =, = in (v) = =± (, ±,) are the critical points = =± ( ±,,) are the other critical points. f = d = at these four points These are the required points at which function is nearest to origin.

5 Find the shortest distance from the origin to the curve = 5 We are to find the minimum value of f = d = + Consider subject to the condition F = + + λ( ) = + λ(+ 8 ) F = + λ(8+ ) For critical points + λ( + ) =.. (i) + λ(+ 7 ) =... (ii) = =.. (iii) (i) ( + λ) + λ= λ (ii) λ+ (+ 7 λ) = + 7λ = λ λ + 7λ = = 6 λ = ( + λ)(+ 7 λ) + λ λ 6λ = + λ+ 7λ + 7λ 9λ 8λ = ( λ )(9λ + ) = λ =, 9 λ = = = Putting this value of in equation (iii) we have ( ) + 8( ) + 7 = = 5 5 = 5 which gives imaginar values of. ( ) 9 λ = 9 = = = = Putting in (iii), we have + 8( ) + 7( ) = = = = 5 = 5 & = 5 = 5 = 5 =± 5 The critical points are ( 5, 5 ) & ( 5, 5) Shortest distance = d = 5 d ± ± = 5 ( 5, 5).

6 Find a point ( z,, ) on the sphere + + z = which is farthest from the point (,,). We are to maimize f( z,, ) = ( ) + ( ) + ( z ) subject to the condition + + z = Let F = ( ) + ( ) + ( z ) + λ( + + z ) For critical points = ( ) + λ= F = ( ) + λ = Fz = ( z ) + λz = and + + z = + λ=.. (i) + λ = (ii) z + λz =.... (iii) + + z =.. (iv) =, + λ = + λ, z = + λ Putting in (iv) + + = + λ + λ + λ = ( + λ) λ + =± λ = ± =, =, z = ± ± ± Clearl,, is the point which is farthest from (,,). 6 Find the etreme values of z = 6, provided & satisf + =. Define F = 6 + λ( + ) For critical points, we have = + λ=. (i) F = + λ =. (ii) and + =... (iii) From (i) and (ii) we have =, = λ λ 5 Putting these values in (iii) we get λ =± 5 λ = = = & = =

7 5 λ = = = 5 5 & = = 5 ( ), 5 5 &, are the critical points. 5 5 A= = λ, B= F =, C= F = λ 5 B AC = λ = ± = 5< F is maimum or minimum at the critical points. Now at And at, 5 5, 5 5 The function is min. at, we have A= 5 >, we have A= 5 < 5, 5 5 and ma. at, Find the critical point of f(, ) = Where =. Test for maima and minima. Define F = λ( ) For critical points, we have = λ=.... (i) F = + + λ = (ii) and =. (iii) From (i) λ = From (ii) λ = + + = + + = = But from (iii) = + ( + ) = = ( + ) + ( + ) = ( + )(+ ) = Either = or = If =, we get = from (iii) (, ) is a critical point and λ = in this case. 9 7 If =, we get = i.e. = 6 6 7, 6 is the other critical point and λ = in this. case.

8 Now A= = + λ, B= F =, C= F = = + = B AC ( λ) 8λ λ = B AC = > f is neither maimum nor minimum at (,). λ = B AC = 8 = + = < and A= + λ = + = > 7, is the point of minimum value. 6 Find the critical points of z = + when + = 6, Also test for maima and minima. Define F = + + λ( + 6 ) For critical points we have F = + λ 6λ=... (i) = + λ 6λ =.. (ii) F and + 6 = (iii) from (i) λ = 6 from (ii) λ = 6 = 6 6 ( 6 ) = ( 6 ) = ( ) ( ) = + = ( ) + ( )( + ) = ( )(+ + ) = Either = or + + = If = then (iii) becomes + 6 = 6 = ( ) = =, =, = & =, = (,) & (,) are the critical points. At (,), λ = = 6 6 = = = = = 6 () 6 And at (,), λ = 8

9 A= = + 6λ B= F = 6λ C= F = + 6λ At (,), we have A =, B =, C = And B AC = Consider z= zhh (, ) z(,) = h + h = h (,) is the point of minimum value. At (,), we have A = + 6 () = B = 6 = C = + 6 () = and B AC = 6 < and A= < (,) is a point of maimum value. 9 Find the points in the plane + z = 5 nearest to the origin. We are to minimize f = d = + + z subject to + z 5=. Define F = + + z + λ( + z 5) = + λ =... (i) F = + λ =... (ii) Fz = z λ =..(iii) and + z 5=..... (iv) = λ, λ λ =, z = from (i), (ii) & (iii) resp. (iv) becomes 9λ λ λ 5= λ + 9λ + λ = 5 λ = = =, =, z = ,, is the critical point. 7 A= F =, B= F =, C= F = B AC = < and A= > 55 5 F is relative minimum at,, so this is the required 7 point.