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

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Guys, does anyone know the answer?

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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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## Solved find the dimensions of the closed rectangular box

Answer to Solved find the dimensions of the closed rectangular box

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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.

Guys, does anyone know the answer?