Question

Show that the rectangular solid of maximum volume that can be inscribed in a given sphere is
a cube.

Answer 1

Step-by-step explanation:

the solution is in the above photo

Answer 2

Answer: 8r³/3√3

Step-by-step explanation:

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

∂Vb/∂x = ∂8.x.y.√r²−x²−y²/∂x

∂Vb/∂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,

∂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.

                                                             #SPJ2