Question

Question #5e9cc

Answer 1

The limiting size of cube is the space diagonal from a lower vertex to an upper vertex. We find this using Pythagoras' theorem.

First, the face diagonal is found

`fd = sqrt(a^2+a^2) = sqrt(2)a` where a is the side length of the cube.

Then the space diagonal is

`sd = sqrt((fd)^2+a^2) = sqrt((sqrt(2)a)^2+a^2) = sqrt(3)a`

The biggest cube that can therefore be cut from the sphere will have a space diagonal of 5 inches, so

`5 = sqrt(3)a implies a = 5/(sqrt(3))` inches

Volume of cube = `a^3 = (5/(sqrt(3)))^3 = ((125)/(3sqrt(3))) = 24.056 "in"^3`

Answer 2

`V = 8 r^3/(3 sqrt(3)) = 24.0563`

Explanation:

Maximum `V = 8xyz`

subjected to

`x^2+y^2+z^2- r^2 = 0`

Using Lagrangian multipliers

`L(x,y,z,lambda)= 8 xyz+lambda (x^2+y^2+z^2-r^2)`

Stationarity condition

`grad L(x,y,z,lambda) = vec 0`

# { (2 lambda x + 8y z = 0), (2 lambda y + 8x z = 0), (2 lambda z + 8x y=0), (x^2 + y^2 + z^2 -r^2= 0) :}#

Solving for `x,y,z,lambda`

`x = r/sqrt[3], y = r/sqrt[3], z = r/sqrt[3], lambda =- (4r)/(2 sqrt[3])`

so `V = 8 r^3/(3 sqrt(3)) =24.0563`