Explanation Given: The function f ( m , b ) = 3 b 2 + 11515856 m 2 + 11752 b m − 5358 b − 105129 m + 240.832 . Critical point is, ( m , b ) = ( 0.01 , − 17.81 ) Formula Used: Let f ( x , y ) is a function whose all partial derivatives f x , f y , f x x , f y y , f x y exist. If a and b be critical point then, f x ( a , b ) = 0 and f y ( a , b ) = 0 . Take a function, M = f x x ( a , b ) . f y y ( a , b ) − [ f x y ( a , b ) ] 2 If M > 0 and f x x ( a , b ) > 0 , then f has minimum at ( a , b ) . Proof: Consider the function, f ( m , b ) = 3 b 2 + 11515856 m 2 + 11752 b m − 5358 b − 105129 m + 240.832 Partially differentiate f ( m , b ) with respect to m , ∂ f ∂ m ( f ( m , b ) ) = 23031712 m + 11752 b − 105129 f m ( m , b ) = 23031712 m + 11752 b − 105129 Partially differentiate f ( m , b ) with respect to b , ∂ f ∂ b f ( m , b ) = 6 b + 11752 m − 5358 f b ( m , b ) = 6 b + 11752 m − 5358 Partially differentiate f m ( m , b ) with respect to m , ∂ f ∂ m ( f m ( m , b ) ) = ∂

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

11th Edition

ISBN: 9781323751671

Author: Lial

Publisher: PEARSON C

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Chapter 14, Problem 1CE

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To prove: The critical points m=0.01 and b=−17.81 corresponds to a minimum of the function,f(m,b)=3b2+11515856m2+11752bm−5358b−105129m+240.832.

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Chapter 14, Problem 54RE

Chapter 14, Problem 2CE

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Chapter 14 Solutions

MATH W/APPLICAT.W/NOTES GDE +ACCESS CODE

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Checkout Lines According to...Ch. 14.1 - Prob. 40E Ch. 14.1 - Prob. 41E Ch. 14.1 - Prob. 42E Ch. 14.1 - Prob. 43E Ch. 14.1 - Prob. 44E Ch. 14.1 - Prob. 45E Ch. 14.1 - Prob. 46E Ch. 14.1 - Prob. 47E Ch. 14.1 - Prob. 48E Ch. 14.2 - Prob. 1CP Ch. 14.2 - Prob. 2CP Ch. 14.2 - Prob. 3CP Ch. 14.2 - Prob. 4CP Ch. 14.2 - Prob. 5CP Ch. 14.2 - Prob. 6CP Ch. 14.2 - Prob. 7CP Ch. 14.2 - Prob. 8CP Ch. 14.2 - Prob. 9CP Ch. 14.2 - Prob. 1E Ch. 14.2 - For each of the given functions, find the...Ch. 14.2 - Prob. 3E Ch. 14.2 - Prob. 4E Ch. 14.2 - Prob. 5E Ch. 14.2 - Prob. 6E Ch. 14.2 - Prob. 7E Ch. 14.2 - Prob. 8E Ch. 14.2 - Prob. 9E Ch. 14.2 - Prob. 10E Ch. 14.2 - Prob. 11E Ch. 14.2 - Prob. 12E Ch. 14.2 - Prob. 13E Ch. 14.2 - Prob. 14E Ch. 14.2 - Prob. 15E Ch. 14.2 - Prob. 16E Ch. 14.2 - Prob. 17E Ch. 14.2 - Prob. 18E Ch. 14.2 - Prob. 19E Ch. 14.2 - Prob. 20E Ch. 14.2 - Prob. 21E Ch. 14.2 - Prob. 22E Ch. 14.2 - Prob. 23E Ch. 14.2 - Prob. 24E Ch. 14.2 - Prob. 25E Ch. 14.2 - Prob. 26E Ch. 14.2 - Prob. 27E Ch. 14.2 - Prob. 28E Ch. 14.2 - Prob. 29E Ch. 14.2 - Prob. 30E Ch. 14.2 - Prob. 31E Ch. 14.2 - Prob. 32E Ch. 14.2 - Prob. 33E Ch. 14.2 - Prob. 34E Ch. 14.2 - Prob. 35E Ch. 14.2 - Prob. 36E Ch. 14.2 - Prob. 37E Ch. 14.2 - Find fx,fy,fz, and fyz for the given functions. 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To prove: The critical points m = 0.01 and b = − 17.81 corresponds to a minimum of the function, f ( m , b ) = 3 b 2 + 11515856 m 2 + 11752 b m − 5358 b − 105129 m + 240.832 .

Solution Summary: The author explains that the critical points m and b correspond to a minimum of the function, f(m,b).

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To prove: The critical points m=0.01 and b=−17.81 corresponds to a minimum of the function,f(m,b)=3b2+11515856m2+11752bm−5358b−105129m+240.832.

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Chapter 14 Solutions