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

To find the maximal volume of a rectangular box inscribed in a sphere, we can use the method of optimization through calculus. Let’s consider that the sphere has a radius r. We need to inscribe a rectangular box such that the corners of the box touch the surface of the sphere.

Let the dimensions of the box be x, y, and z. The volume V of the box can be expressed as:

V = x y z

The relationship between the box dimensions and the radius of the sphere is given by the equation of a sphere:

x² + y² + z² = r²

To maximize the volume, we can use the method of Lagrange multipliers or substitute one of the variables in terms of the others. A convenient approach is to express z in terms of x and y:

z = sqrt(r² – x² – y²)

Substituting this into the volume formula gives:

V = x y sqrt(r² – x² – y²)

We can then take partial derivatives of V with respect to x and y, set them equal to zero to find critical points, and analyze these points to determine where the volume is maximized.

It turns out that for a rectangular box inscribed in a sphere, the maximum volume occurs when the box is a cube. Hence, if the radius of the sphere is r, the dimensions of the maximal inscribed cube will be:

x = y = z = r/√3

This gives the maximal volume:

V = (r/√3)³ = r³ / 3√3

In conclusion, the maximal volume of a rectangular box inscribed in a sphere is achieved when the box is a cube, and is expressed as V = r³ / 3√3.