{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chemical Engineering Hand Written_Notes_Part_82

# Chemical Engineering Hand Written_Notes_Part_82 - 166 5...

This preview shows pages 1–2. Sign up to view the full content.

166 5. OPTIMIZATION AND RELATED NUMERICAL SCHEMES Theorem 11 . If F ( z ) is continuous and di ff erentiable and has an extreme (or stationary) point (i.e. maximum or minimum ) point at z = z , then (2.1) F ( z ) = ∂F ∂z 1 ∂F ∂z 2 .............. ∂F ∂z N ¸ T z = z = 0 . Proof: Suppose z = z is a minimum point and one of the partial deriva- tives, say the k th one, does not vanish at z = z , then by Taylor’s theorem (2.2) F ( z + z ) = F ( z ) + N X i =1 ∂F ∂z i ( z ) z i + R 2 ( z , z ) (2.3) i.e. F ( z + z ) F ( z ) = z k ∂F ∂z i ( z ) + R 2 ( z , z ) Since R 2 ( z , z ) is of order ( z i ) 2 , the terms of order z i will dominate over the higher order terms for su ciently small z . Thus, sign of F ( z + z ) F ( z ) is decided by sign of z k ∂F ∂z k ( z ) Suppose, (2.4) ∂F ∂z k ( z ) > 0 then, choosing z k < 0 implies (2.5) F ( z + z ) F ( z ) < 0 F ( z + z ) < F ( z ) and F ( z ) can be further reduced by reducing z k . This contradicts the assump- tion that z = z is a minimum point. Similarly, if

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}