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Section 12.8 Lagrange Multipliers The problem is to find the extreme values of a given function f ( x , y , z ) subject to the constraint g ( x , y , z ) k . The method of Lagrange multipliers is to solve f g for some constant . Intuitive explanation: g ( x , y , z ) k is a level surface of g. Suppose that f ( x , y , z ) has an extreme value at some point (a, b, c) on this surface. Then for any curve on this surface with vector equation r (t ) with r ( t 0 ) a , b , c , the function f ( x ( t ), y ( t ), z ( t )) has an extreme value at t 0 . So But d d dt f ( x ( t ), y ( t ), z ( t )) 0 at t 0 . f ( x ( t ), y ( t ), z ( t )) f r ( t ) dt . So f ( a , b , c ) r ( t 0 ) 0 . We have shown that the gradient of f at (a, b, c) is orthogonal to every curve on g ( x , y , z ) k that passes through (a, b, c). So the gradient of f at (a, b, c) is orthogonal to the tangent plane to the surface at this point and must be parallel to the gradient of g at this point. Examples: Find the extreme values of the given function subject to the given constraint. 1. Pg 788 problem 1 in Stewart 2 f ( x, y ) x y 2 subject to 2 x y 2 1. 2. pg 788 problem 2 f ( x, y ) 2 x y subject to 2 x 4y 2 1 3. pg 788 problem 8 2 2 2 f ( x, y, z) x y z subject to 2 2 2 x y z 1