Question

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

Answer 1

Answer:

Step-by-step explanation:

Let the rectangle have (x,y,z) as vertex in positive octant.  The rectangular box has to be necessarily symmetrical about all the three axes.

Then the sides of the box would be

$2x,2y,2z$

Volume = $8xyz$

Maximize volume subject to

$x^2+y^2+z^2 =1$

i.e. $g(x,y,z) = x^2+y^2+z^2 -1=0$

Use Lagrangian multipliers , we have

$∇f(x,y,z)=λ∇g(x,y,z)$at the maximum

$∇f(x,y,z)=λ∇g(x,y,z) =(8yz,8xz,8xy)\\∇g(x,y,z)=(2x,2y,2z)$

$8yz=2λx8xz=2λy8xy=2λz$

Dividing we get

$\frac{y}{x} =\frac{x}{y} \\x^2=y^2$

Similarly $y^2=z^2$

Thus we get $3x^2 =1\\x = \frac{1}{\sqrt{3} }$

Hence dimensions are

(2x,2y,2z)

So dimensions are

$\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } ,\frac{2}{\sqrt{3} } )$