lecture12

# lecture12 - MATH 100 Lecture 12 Lagrange multiplier Example...

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2006 Fall MATH 100 Lecture 12 1 MATH 100 Lecture 12 Lagrange multiplier Example: Minimize the surface of a open box subject to fixed volume 32 subject to minimize = + + = xyz xz yz xy S Two approaches First approach: solve the constraint for one variable z=h(x,y) , substitute into s(x,y,z) and get f(x,y,h(x,y)), and compute extremum of f(x,y,h(x,y))

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2006 Fall MATH 100 Lecture 12 2 Constrained minimization in 2-space and 3-space ( ) () 0 , s.t. , min = y x g y x f 0 , , s.t. , , min = z y x g z y x f ( ) ( ) minimum relative d constraine , , and , solution the call We 0 0 0 0 0 z y x y x MATH 100 Lecture 12 Lagrange multiplier
2006 Fall MATH 100 Lecture 12 3 Second approach: Lagrange multiplier () 00 Theorem: (The constrained-extremum principle of 2-variables) If 0, then relative extremum of subject to 0 occurs at points , satisfying , , g f g xy fxy g λ ∇≠ = ∇= for some Lagrange multiplier ( ) ( ) () () Proof: Parametrization ,0 , ', ' ( ) 0 , gxy x xt y yt gx ty t g r t t =⇔= = ⇒∇ ⋅ =∇ ⋅ = ∀ K MATH 100 Lecture 12 Lagrange multiplier

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2006 Fall MATH 100 Lecture 12 4 ( ) ( ) ( ) ( ) ( ) t F t y t x f y x f t = = , , , of In term () () () ( ) ( ) 0 0 0 00 If has relative extremum at , ' 0 , , ' , ' 0 so ' , ' ( ) 0 xy tt F t t that Ft d fx t yt f xyxt f xyyt dt t y t f r t = = = ⎡⎤
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## This note was uploaded on 09/17/2011 for the course MATH 100 taught by Professor Qt during the Fall '09 term at HKUST.

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lecture12 - MATH 100 Lecture 12 Lagrange multiplier Example...

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