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mthsc206-fall-2010-notes-15.8

mthsc206-fall-2010-notes-15.8 - g x y z = k 2 Evaluate f at...

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15.8 Lagrange Multipliers Theorem: Let f and g be functions of 2( or 3) variables with continuous first partial derivatives on some open set containing the constraint curve (or surface) g ( x, y ) = k (or g ( x, y, z ) = k ) and assume g = 0 at any point on this curve (or surface). If f has a constrained relative extremum, then this extremum occurs at a point P 0 on the constraint curve (or surface) at which the gradient vectors f ( P 0 ) and g ( P 0 ) are parallel; i.e., there is some number λ such that f ( P 0 ) = λ g ( P 0 ) . Method of Lagrange Multipliers: To find the maximum and minimum values of f ( x, y, z ) subject to the constraint g ( x, y, z ) = k (assuming that these extreme values exist and g = 0 on the surface g ( x, y, z ) = k ): 1. Find all values of x , y , z , and λ such that f ( x, y, z ) = λ g ( x, y, z ) AND
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Unformatted text preview: g ( x, y, z ) = k. 2. Evaluate f at all of the points ( x, y, z ) that result from step 1. The largest of these values is the maximum of f subject to the constraint; the smallest of these values is the minimum of f subject to the constraint. Example 15.8.1 Redo fnding the absolute extrema oF f ( x, y ) = x 2 + 2 y 2-x on the region x 2 + y 2 ≤ 4 using the Method oF Lagrange Multipliers. Example 15.8.2 ±ind the extrema oF f ( x, y ) = xy iF ( x, y ) is restricted to the ellipse 4 x 2 + y 2 = 4 . Example 15.8.3 IF f ( x, y, z ) = 4 x 2 + y 2 + 5 z 2 , fnd the point on the plane 2 x + 3 y + 4 z = 12 at which f ( x, y, z ) has its least value. 33...
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