Explanation Given information: The equation of a sphere x 2 + y 2 + z 2 = 4 . Calculation: Let x denote the length, y denotes the breadth, and z denote the height of the rectangular box. The volume of the rectangular box inside thesphere in terms of coordinates x , y & z shall be, V b = 2 x ⋅ 2 y ⋅ 2 z = 8 x y z . The maximum volume 8 x y z of a rectangular box is subject to the constraint x 2 + y 2 + z 2 = 4 . Therefore,the objective function is f x , y , z = 8 x y z ,for the volume of the box to be non-zero. we must have x > 0 , y > 0 , z > 0 . Write the constraint equation x 2 + y 2 + z 2 = 4 as x 2 + y 2 + z 2 − 4 = 0 . Let g x , y , z = x 2 + y 2 + z 2 − 4 . Formthe Lagrange function as, L x , y , z = f x , y , z − λ g x , y , z = x y z − λ x 2 + y 2 + z 2 − 4 = x y z − λ x 2 − λ y 2 − λ z 2 + 4 λ Compute the first partial derivatives of L x . L x = ∂ L ∂ x = ∂ x y z − λ x 2 − λ y 2 − λ z 2 + 4 λ ∂ x = y z − 2 λ x Therefore, L x = y z − 2 λ x . Compute the first partial derivatives of L y . L y = ∂ L ∂ y = ∂ x y z − λ x 2 − λ y 2 − λ z 2 + 4 λ ∂ y = x z − 2 λ y Therefore, L y = x z − 2 λ y . Compute the first partial derivatives of L z . L z = ∂ L ∂ z = ∂ x y z − λ x 2 − λ y 2 − λ z 2 + 4 λ ∂ z = x y − 2 λ z Therefore, L z = x y − 2 λ z . Compute the first partial derivatives of L λ . L λ = ∂ L ∂ λ = ∂ x y z − λ x 2 − λ y 2 − λ z 2 + 4 λ ∂ λ = − x 2 − y 2 − z 2 + 4 Therefore, L λ = − x 2 − y 2 − z 2 + 4 . Solve the system of equation L x = 0 , L y = 0 , L z = 0 & L λ = 0 , that is, y z − 2 λ x = 0 ... 1 x z − 2 λ y = 0 ... 2 x y − 2 λ z = 0 ... 3 − x 2 − y 2 − z 2 + 4 = 0 ... 4 Simplify equation 1 for λ . y z − 2 λ x = 0 λ = y z 2 x

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Chapter 6.5, Problem 40E

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To find:The maximum volume of a rectangular box that can be contained inside the sphere x2+y2+z2=4 .

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Chapter 6.5, Problem 39E

Chapter 6.5, Problem 41E

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Chapter 6 Solutions

CALCULUS+ITS...,EXP.(LL)-W/CODE NVCC

Ch. 6.1 - 2. . Ch. 6.1 - Forf(x,y)=x23xy,find(0,2),f(2,3),andf(10,5).Ch. 6.1 - Prob. 3E Ch. 6.1 - 3. . Ch. 6.1 - 6. . Ch. 6.1 - Forf(x,y)=Inx+y3,findf(e,2),f(e2,4),andf(e3,5).Ch. 6.1 - 8. . Ch. 6.1 - Forf(x,y,z)=x2y2+z2,findf(1,2,3)andf(2,1,3).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 maximum volume of a rectangular box that can be contained inside the sphere x 2 + y 2 + z 2 = 4 .

Solution Summary: The author calculates the maximum volume of a rectangular box that can be contained inside the sphere.

To find:The maximum volume of a rectangular box that can be contained inside the sphere x2+y2+z2=4 .

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Chapter 6 Solutions