Explanation Given: x + 1 2 y + 1 3 z = 1 Let P = ( a , b ) be a critical point of f ( x , y ) . Let the second order derivatives f x x , f y y , f x y be continuous near P . Let D be the discriminant of the function f ( x , y ) and is defined as D ( x , y ) = f x x f y y − f x y 2 (a) If D > 0 , then f ( a , b ) is a local minimum or local maximum. (b) If D < 0 , then ( a , b ) is a saddle point. (c) If D = 0 , then the test is inconclusive. (d) If D > 0 and f x x ( a , b ) > 0 then f ( a , b ) is a local minimum. (a) If D > 0 and f x x ( a , b ) < 0 then f ( a , b ) is a local maximum. Calculation: x + 1 2 y + 1 3 z = 1 1 3 z = 1 − x − 1 2 y z = 3 − 3 x − 3 2 y The volume of the box V = x y z V = x y ( 3 − 3 x − 3 2 y ) = 3 x y − 3 x 2 y − 3 2 x y 2 V x = 3 y − 6 x y − 3 2 y 2 V y = 3 x − 3 x 2 − 3 2 2 x y = 3 x − 3 x 2 − 3 x y Putting V x = 0 3 y − 6 x y − 3 2 y 2 = 0 3 y ( 1 − 2 x − 1 2 y ) = 0 ⇒ y = 0 , 1 − 2 x − 1 2 y = 0 2 x = 1 − 1 2 y x = 1 2 − 1 4 y ... ( 1 ) Putting V y = 0 3 x − 3 x 2 − 3 x y = 0 3 x ( 1 − x − y ) = 0 ⇒ x = 0 , 1 − x − y = 0...

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)

3rd Edition

ISBN: 9781319019846

Author: Jon Rogawski, Colin Adams

Publisher: W. H. Freeman

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Chapter 14.7, Problem 46E

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The maximum volume of the box inscribed in the tetrahedron bounded by the coordinate planes and the plane $x + \frac{1}{2} y + \frac{1}{3} z = 1$

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Chapter 14.7, Problem 45E

Chapter 14.7, Problem 47E

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Chapter 14 Solutions

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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The maximum volume of the box inscribed in the tetrahedron bounded by the coordinate planes and the plane x + 1 2 y + 1 3 z = 1

The maximum volume of the box inscribed in the tetrahedron bounded by the coordinate planes and the plane x+12y+13z=1

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Chapter 14 Solutions

The maximum volume of the box inscribed in the tetrahedron bounded by the coordinate planes and the plane $x + \frac{1}{2} y + \frac{1}{3} z = 1$