Explanation Let the equation of a plane be x a + y b + z c = 1 . The smallest volume which cuts from the first octant is tetrahedron and is given by V = a b c 6 . Substitute ( x , y , z ) = ( 1 , 2 , 3 ) in the equation x a + y b + z c = 1 , 1 a + 2 b + 3 c = 1 . (1) Let V = a b c 6 . (2) In the above equation take c has a function of ( a , b ) . Take the partial derivative of V ( a , b ) with respect to a and obtain V a . ∂ V ∂ a = ∂ ∂ a ( a b c 6 ) = b 6 ∂ ∂ a ( a c ) = b 6 [ c ( 1 ) + a ∂ c ∂ a ] = b c 6 + a b 6 ∂ c ∂ a Hence, V a = b c 6 + a b 6 ∂ c ∂ a . Take the partial derivative of V ( a , b ) with respect to b and obtain V b . ∂ V ∂ b = ∂ ∂ b ( a b c 6 ) = a 6 ∂ ∂ b ( b c ) = a 6 [ c ( 1 ) + b ∂ c ∂ a ] = a c 6 + a b 6 ∂ c ∂ a Thus, V b = a c 6 + a b 6 ∂ c ∂ a . Differentiate the equation (1) on both sides with respect to a ∂ ∂ a ( 1 a + 2 b + 3 c ) = ∂ ∂ a ( 1 ) ∂ ∂ a ( 1 a ) + ∂ ∂ a ( 2 b ) + ∂ ∂ a ( 3 c ) = 0 ( − 1 a 2 ) + 0 + 3 ( − 1 c 2 ) ∂ c ∂ a = 0 ∂ c ∂ a = − c 2 3 a 2 Differentiate the equation (1) on both sides with respect to b

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Chapter 11.7, Problem 48E

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To find: The equation of the plane that passes through the point (1,2,3) and cuts off the smallest volume from the first octant.

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Chapter 11.7, Problem 47E

Chapter 11.8, Problem 20E

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Chapter 11 Solutions

Essential Calculus: Early Transcendentals

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The equation of the plane that passes through the point ( 1 , 2 , 3 ) and cuts off the smallest volume from the first octant.

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To find: The equation of the plane that passes through the point (1,2,3) and cuts off the smallest volume from the first octant.

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Chapter 11 Solutions