Explanation Given Information: The provided function is, f ( x , y ) = 1 , 000 + 5 x − 4 y − 3 x y Formula Used: The partial derivatives of f with respect to x is the derivative of f with respect to x , when all other variables are treated as constant and is denoted as ∂ f ∂ x . The second order partial derivative of f with respect to x is denoted as, ∂ 2 f ∂ x 2 = ∂ ∂ x ( ∂ f ∂ x ) The derivative formula, d d x ( x n ) = n x n − 1 And, d d x ( k ⋅ f ( x ) ) = k d d x ( f ( x ) ) And, d d x ( constant ) = 0 Where, n is any real number and k is a constant. Calculation: Consider the provided function, f ( x , y ) = 1 , 000 + 5 x − 4 y − 3 x y To calculate the value of ∂ f ∂ x , treat 4 y and 3 y as a constant. Thus, ∂ f ∂ x = ∂ ∂ x ( 1 , 000 + 5 x − 4 y − 3 x y ) = ∂ ∂ x ( 1 , 000 ) + ∂ ∂ x ( 5 x ) − ∂ ∂ x ( 4 y ) − ∂ ∂ x ( 3 x y ) = 0 + 5 + 0 − 3 y = 5 − 3 y Again, find the partial derivative of above equation with respect to x , to find ∂ 2 f ∂ x 2 , ∂ 2 f ∂ x 2 = ∂ ∂ x ( ∂ f ∂ x ) = ∂ ∂ x ( 5 − 3 y ) = 0 + 0 = 0 And, for the value of ∂ 2 f ∂ x 2 | ( 1 , − 1 ) , substitute 1 for x and − 1 for y in the value obtained for ∂ 2 f ∂ x 2 , ∂ 2 f ∂ x 2 | ( 1 , − 1 ) = 0 Now, to calculate the value of ∂ f ∂ y , treat 5 x and 3 x as a constant. Thus, ∂ f ∂ y = ∂ ∂ y ( 1 , 000 + 5 x − 4 y − 3 x y ) = ∂ ∂ y ( 1 , 000 ) + ∂ ∂ y ( 5 x ) − ∂ ∂ y ( 4 y ) − ∂ ∂ y ( 3 x y ) = 0 − 0 − 4 − 3 x = − 3 x − 4 Again, find the partial derivative of above equation with respect to y , to find ∂ 2 f ∂ y 2 , ∂ 2 f ∂ y 2 = ∂ ∂ y ( ∂ f ∂ y <

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

7th Edition

ISBN: 9781337291293

Author: Waner, Stefan; Costenoble, Steven

Publisher: Brooks Cole

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Question

Chapter 8.2, Problem 22E

To determine

To calculate: The values of ∂2f∂x2,∂2f∂y2,∂2f∂x∂y and ∂2f∂y∂x where f(x,y)=1,000+5x−4y−3xy. Also, evaluate them all at (1,−1) if possible.

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Chapter 8.2, Problem 21E

Chapter 8.2, Problem 23E

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Chapter 8 Solutions

Student Solutions Manual for Waner/Costenoble's Applied Calculus, 7th

Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 14, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 6E Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - For each function in Exercises 5-8, evaluate (a)...Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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The values of ∂ 2 f ∂ x 2 , ∂ 2 f ∂ y 2 , ∂ 2 f ∂ x ∂ y and ∂ 2 f ∂ y ∂ x where f ( x , y ) = 1 , 000 + 5 x − 4 y − 3 x y . Also, evaluate them all at ( 1 , − 1 ) if possible.

Solution Summary: The author explains how to calculate the partial derivatives of f with respect to x.

To calculate: The values of ∂2f∂x2,∂2f∂y2,∂2f∂x∂y and ∂2f∂y∂x where f(x,y)=1,000+5x−4y−3xy. Also, evaluate them all at (1,−1) if possible.

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Chapter 8 Solutions