Question

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

Answer 1

Step 1: Find the expression for volume.

The radius of a sphere is around a fixed point can be given as,

${\text{f(x, y, z)=x}}^2+y^2+z^2$

(Assume the iteration: ${\text{g(x, y, z)=x}}^2+y^2+z^2-1=0$ …..$\left(1\right)$

Let $2x,2y$ and $\text{2z}$ be the length, width, and height respectively of the rectangular box.

the volume of the box is,

$$\begin{gathered} V=2x\times 2y\times 2z \\ \Rightarrow V=8xyz \end{gathered}$$

Step 2. Find the dimensions of the closed rectangular box.

Now, use the Lagrange multiplier technique to find the dimension of the rectangular box:

$$\begin{gathered} \nabla V=\lambda \nabla g \\ \Rightarrow (yz,xz,xy)=\lambda (2x,2y,2z)...\left(2\right) \end{gathered}$$

$\left(\begin{array}{cc}yz=& \lambda x\\ xz=& \lambda y\\ xy=& \lambda z\end{array}\right)\text{solving}$

From the first two equations, find $\lambda$ and equate them:

$$\begin{gathered} \lambda =\frac{yz}{x} \\ \lambda =\frac{xz}{y} \\ So,\frac{yz}{x}=\frac{xz}{y} \\ \Rightarrow y^2=x^2 \end{gathered}$$

Take the square root of both the side:

$y=\pm x$

Similarly, $z=\pm x$

From the $\left(1\right)$ equation:

$$\begin{gathered} x^2+y^2+z^2-1=0 \\ \Rightarrow x^2+y^2+z^2=1 \end{gathered}$$

Rewrite the above equation as $x,y$ and $z$ are similar:

$\begin{array}{rcl}{x}^2+x^2+x^2& =& 1\\ & \Rightarrow & 3x^2=1\\ & \Rightarrow & x=\pm \sqrt{\frac{1}{3}}\end{array}$

Thus, $\begin{array}{rcl}2x& =& \pm 2\sqrt{\frac{1}{3}}\end{array}$

Since $2x,2y$ and $2z$ are similar, the dimensions of a closed rectangular box are,

$\pm 2\sqrt{\frac{1}{3}},\pm 2\sqrt{\frac{1}{3}},\pm 2\sqrt{\frac{1}{3}}$.

Hence, the dimensions are $\pm 2\sqrt{\frac{1}{3}},\pm 2\sqrt{\frac{1}{3}},\pm 2\sqrt{\frac{1}{3}}$.