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    Find the dimensions of the closed rectangular box with maxim

    Find step-by-step Calculus solutions and your answer to the following textbook question: Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere..

    Question

    Find the dimensions of the closed rectangular box with maximum volume that can be inscribed in the unit sphere.

    Explanations

    Verified Explanation A Explanation B Restriction:

    g(x,y,z)=x^2+y^2+z^2-1=0

    g(x,y,z)=x 2 +y 2 +z 2 −1=0 V=xyz V=xyz

    \nabla V=yzi+xzj+xyk

    ∇V=yzi+xzj+xyk

    \nabla g=2xi+2yj+2zk

    ∇g=2xi+2yj+2zk.

    We want to know x,y,z that maximaze V.

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    How can you find the maximal volume of a rectangular box inscribed in a sphere?

    How can you find the maximal volume of a rectangular box inscribed in a sphere - The maximal volume of the rectangle inside the sphere of radius r and that volume is 8r3/3√3.

    How can you find the maximal volume of a rectangular box inscribed in a sphere?

    Solution:

    To find the maximum volume of the rectangle inscribed in a sphere, we begin with the general equation of the sphere of radius r in terms of coordinates x, y and z, which is

    x2 + y2 + z2 = r2 --- (1)

    The volume of the rectangular box inside the sphere in terms of coordinates x, y, and z shall be

    Vb = 2x.2y.2z = 8xyz --- (2)

    To find the maximal volume of the box one has to differentiate partially equation (2) w.r.t the variables in which volume is expressed.

    One shall substitute the value of z in terms of x and y in equation 2 to simplify the maximization process. So we get

    z = √r2 - x2 - y2

    Therefore the volume of the rectangular box can be expressed as:

    Vb = 8.x.y. √r2 - x2 - y2 --- (3)

    Partially differentiating the above w.r.t. x

    ∂ V b ∂ x ∂Vb∂x = ∂ 8. x . y . √ r 2 − x 2 − y 2 ∂ x ∂8.x.y.r2−x2−y2∂x ∂ V b ∂ x ∂Vb∂x = √ r 2 − x 2 − y 2 ∂ ( 8 x y ) ∂ x + 8 x y ∂ √ r 2 − x 2 − y 2 ∂ x

    r2−x2−y2∂(8xy)∂x+8xy∂r2−x2−y2∂x

    = √r2 - x2 - y2 × (8y) + (8xy)(-2x)(1/2)(r2 - x2 - y2)-1/2

    = 8y × √r2 - x2 - y2 - 8x2y/(√r2 - x2 - y2)

    = 8y(r2 - x2 - y2) - 8x2y/(√r2 - x2 - y2)

    = 8y(r2 - x2 - y2 - x2)/(√r2 - x2 - y2)

    = 8y((r2 -2x2 - y2)/(√r2 - x2 - y2) --- (4)

    Similarly differentiating equation (3) partially w.r.t. y we get,

    ∂ V b ∂ x ∂Vb∂x

    = 8x((r2 - x2 - 2y2)/(√r2 - x2 - y2) --- (5)

    Equations (4) and (5) are partial derivatives and have to be equated to zero to maximize the volume of the rectangular box.

    Since the denominator of both equations (4) & (5) cannot be zero and x and y cannot be zero the only alternative left is:

    (r2 -2x2 - y2) = 0 ⇒ 2x2 + y2 = r2 --- (6)

    (r2 - x2 - 2y2) = 0 ⇒ x2 + 2y2 = r2 --- (7)

    Multiplying (7) by 2 and subtracting from (6) we get,

    -3y2 = -r2 ⇒ y = r/√3 --- (8)

    Similarly multiplying (6) by 2 and subtracting (7) from it) we get,

    3x2 = r2 ⇒ x = r/√3 --- (9)

    Substituting (8) and (9) in the volume equation (3) we have,

    Vb = 8(r/√3)(r/√3)√(r2 - (r/√3)2 - ( r/√3)2

    Vb = 8(r2/3)√(r2 - r2/3 - r2/3)

    = (8/3)(r2)√3r2 - r2 - r2)/3

    = (8/3)(r2)√r2/3 = 8r3/3√3

    Hence, the maximal volume of a rectangular box inside the sphere is 8r3/3√3.

    How can you find the maximal volume of a rectangular box inscribed in a sphere?

    Summary:

    The maximal volume of the rectangle inside the sphere of radius r and that volume is 8r3/3√3.

    Source : www.cuemath.com

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