Explanation Given: The function is f ( x , y ) = x 2 + 4 y 3 − 6 x y − 1 Calculation: Consider the provided function: f ( x , y ) = x 2 + 4 y 3 − 6 x y − 1 Since the partial derivatives always exist, find all points ( a , b ) such that f x ( a , b ) = 0 f y ( a , b ) = 0 Here, f x = ∂ ∂ x ( x 2 + 4 y 3 − 6 x y − 1 ) = 2 x − 6 y and f y = ∂ ∂ y ( x 2 + 4 y 3 − 6 x y − 1 ) = 12 y 2 − 6 x Now, set each partial derivative equal to 0: 2 x − 6 y = 0 (1) and 12 y 2 − 6 x = 0 …...…... (2) Now, to solve the equations (1) and (2) by elimination, divide the equation (2) by 3 and then add the equations: ( 2 x − 6 y ) + ( 12 y 2 − 6 x ) 3 = 0 2 x − 6 y + 4 y 2 − 2 x = 0 − 2 y ( 3 − 2 y ) = 0 y = 0 , or 1.5 Now, substitute y = 0 in the equation (1): x = 0 Again, substitute y = 1.5 in the equation (1): 4 x − 12 ⋅ ( 1.5 ) = 0 4 x = 12 ⋅ ( 1.5 ) x = 4.5 Therefore, ( 0 , 0 ) and ( 4.5 , 1.5 ) are the solutions of the system. Now, find the various second-order partial derivatives used to find M . f x x = 2 f y y = 24 y f x y = − 6 Now, find M at the point ( 0 , 0 )

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Chapter 14.3, Problem 19E

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To calculate: All points at which the provided functions have any local extrema, the values of any local extrema, and also the saddle points if the function is f(x,y)=x2+4y3−6xy−1

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Chapter 14.3, Problem 18E

Chapter 14.3, Problem 20E

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All points at which the provided functions have any local extrema, the values of any local extrema, and also the saddle points if the function is f ( x , y ) = x 2 + 4 y 3 − 6 x y − 1

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To calculate: All points at which the provided functions have any local extrema, the values of any local extrema, and also the saddle points if the function is f(x,y)=x2+4y3−6xy−1

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Chapter 14 Solutions