Explanation Given Information: The expression is f ( x , y ) = x 3 + y 3 − 3 x − 12 y . Formula used: The D can be calculated by: D = f x x ( a , b ) ⋅ f y y ( a , b ) − [ f x y ( a , b ) ] 2 Where f x x , f y y and f x y are the second partial derivatives. Condition for f to has a maximum at ( a , b ) , if D > 0 and f x x < 0 . Condition for f to has a minimum at ( a , b ) , if D > 0 and f x x > 0 . And f has neither a maximum nor a minimum at ( a , b ) , if D < 0 . Calculation: Consider the expression, f ( x , y ) = x 3 + y 3 − 3 x − 12 y Find f x f x = ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) = 3 x 2 + 0 − 3 − 0 = 3 x 2 − 3 Find f y f y = ∂ ∂ y ( x 3 + y 3 − 3 x − 12 y ) = 0 + 3 y 2 − 0 − 12 = 3 y 2 − 12 Find f x x f x x = ∂ ∂ x [ ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) ] = ∂ ∂ x ( 3 x 2 + 0 − 3 − 0 ) = ∂ ∂ x ( 3 x 2 − 3 ) = 6 x Find f y y f y y = ∂ ∂ y ( 3 y 2 − 12 ) = 6 y Find f x y f x y = ∂ ∂ y [ ∂ ∂ x ( x 3 + y 3 − 3 x − 12 y ) ] = ∂ ∂ y ( 3 x 2 + 0 − 3 − 0 ) = ∂ ∂ y ( 3 x 2 − 3 ) = 0 Solve the system of equations f x = 0 and f y = 0 3 x 2 − 3 = 0 … … ( 1 ) 3 y 2 − 12 = 0 … … ( 2 ) Solve equation ( 1 ) for x , we get 3 x 2 − 3 = 0 3 x 2 = 3 x 2 = 1 x = ± 1 Solve equation ( 2 ) for y , we get 3 y 2 − 12 = 0 3 y 2 = 12 y 2 = 4 y = ± 2 Thus, ( − 1 , − 2 ) and ( 1 , 2 ) are possiblepoints for maximum or minimum

Calculus and Its Applications (11th Edition)

11th Edition

ISBN: 9780321979391

Author: Marvin L. Bittinger, David J. Ellenbogen, Scott J. Surgent

Publisher: PEARSON

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4. Identify at least two mistakes in Francisco's work. Correct the mistakes and complete the problem by using the second derivative test. 2f 2X 2. Find the relative maximum and relative minimum points of f(x) = 2x3 + 3x² - 3, using the First Derivative Test or the Second Derivative Test. bx+ bx 6x +6x=0 12x- af 24 = 0 x=0 108 -2 5. Identify at least three mistakes in Francisco's work. Then sketch the graph of the function and label the local max and local min. 1. Find the equation of the tangent line to the curve y=x-2x3+x-2 at the point (1.-2). Sketch the araph of y=x42x3+x-2 and the tangent line at (1,-2) y' = 4x-6x y' (1) = 4(1) - 667 - 2 = 4(-2)4127-6(-2) 5-8-19-20 =

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Chapter 6 Solutions

Calculus and Its Applications (11th Edition)

Ch. 6.1 - Forf(x,y)=x23xy,find(0,2),f(2,3),andf(10,5).Ch. 6.1 - 2. . Ch. 6.1 - 3. . Ch. 6.1 - 4. . Ch. 6.1 - Forf(x,y)=Inx+y3,findf(e,2),f(e2,4),andf(e3,5).Ch. 6.1 - 6. . Ch. 6.1 - Forf(x,y,z)=x2y2+z2,findf(1,2,3)andf(2,1,3).Ch. 6.1 - 8. . Ch. 6.1 - In Exercises 9-14, determine the domain of each...Ch. 6.1 - In Exercises 9-14, determine the domain of each...

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The relative maximum and minimum values for the equation, f ( x , y ) = x 3 + y 3 − 3 x − 12 y .

Solution Summary: The author calculates the relative maximum and minimum values for the equation f(x,y)=x3+y