Chapter 4, Problem 22P

To determine

The height of the pyramid of minimum volume whose base is a square and whose base triangular faces are all tangents to the sphere.

Explanation of Solution

Given information:

The sphere has radius r. The base of the pyramid is a regular n-gon.

Calculations:

Let the height of the Pyramid for minimum volume be $h$

Radius of the sphere is $r$

Therefore,

  $h=r+x\text{ Where, x= Distance between the center of sphere and cone of pyramid}\text{.}$

Let the side of the square base be $2b$

Area of the square base is $4b^2$

Now from similar triangles rules we get,

  $\begin{array}{l}\frac{b}{x+r}=\frac{r}{\sqrt{x^2-r^2}}\\ \Rightarrow b=\frac{r(x+r)}{\sqrt{x^2-r^2}}\end{array}$

Now volume of the pyramid is,

  $\begin{array}{l}v=\frac{1}{3}Ah=\frac{1}{3}(x+r)\mathrm{.4.}{r}^2\frac{{\left(x+r\right)}^2}{\sqrt{{(x^2-r^2)}^2}}\\ v=\frac{4.r^3}{3}\frac{{(x+r)}^2}{(x-r)}\end{array}$

Now we differentiate to get the minimum value,

  $\begin{array}{l}v\text{'}=\frac{4.r}{3}.\frac{2.(x+r)(x-r)-{(x+r)}^2.1}{{(x-r)}^2}\\ \text{Now equating }v\text{'}=0\text{ ,we get,}\\ v\text{'}=\frac{4.r}{3}.\frac{2.(x+r)(x-r)-{(x+r)}^2.1}{{(x-r)}^2}=0\\ x=3r\\ \text{Earlier we assumed that,}\\ h=x+r\\ \text{Now putting the value of }x\text{ we get,}\\ h=x+r=3r+r\\ h=4r.\end{array}$