What is the maximum volume of an equilateral triangular prism inscribed in a sphere of radius 2? What is the maximum volume of an equilateral triangular prism inscribed in a sphere of radius 2?
Since the volume of an equilateral triangular prism is $\frac{\sqrt3}{4}a^2h$,where $a$ is the side length of the base triangle and $h$ is the height of the prism.How to express this volume in terms of radius of the sphere so that i can differentiate it and equate it to zero.Thanks. 
 A: The triangular prism takes up a certain fraction of the volume of its circumscribed cylinder. Therefore we determine first the maximal cylinder inscribed in the sphere, and bother about the equilateral triangle afterwards.
When the cylinder has height $2x$ its radius is $\rho=\sqrt{4-x^2}$, and  its volume is $V_{\rm cyl}=2x\cdot\pi\rho^2=2\pi x(4-x^2)$. The latter is maximal when $4-3x^2=0$, or $x^2={4\over3}$. It follows that the maximal cylinder has
$$\rho^2=4-x^2={8\over3}\ .$$
The area of the inscribed equilateral triangle then computes to
$$A_\triangle=3\cdot{1\over2}\rho^2\sin(120^\circ)=2\sqrt{3}\ ,$$
so that the maximal prism obtains volume
$$V_{\rm prism}=2x\cdot A_\triangle=8\ .$$
A: Let each side of the equilateral triangular face of the prism be $x$ inscribed in a sphere of radius $R=2$ the inscribed radius of the equilateral triangular base $$=\frac{x}{\sqrt 3}$$
Using geometry of right triangle, the normal distance of the center of equilateral triangular face of prism from the center of sphere $$=\sqrt{R^2-\left(\frac{x}{\sqrt 3}\right)^2}=\frac{\sqrt{3(2)^2-x^2}}{\sqrt 3}=\frac{\sqrt{12-x^2}}{\sqrt 3}$$
then the normal height $(H)$ of the prism is given as
$$H=2\frac{\sqrt{12-x^2}}{\sqrt 3}$$ 
The volume of the inscribed prism $$V=\text{(area of equilateral triangular face)}\times \text{(normal height of prism (H)}$$
$$V=\frac{\sqrt3}{4}x^2\times 2\frac{\sqrt{12-x^2}}{\sqrt 3}$$
$$V=\frac{1}{2}x^2\sqrt{12-x^2}\tag 1$$
Now, differentiating (1) w.r.t. $x$, we get 
$$\frac{dV}{dx}=\frac{1}{2}\left(x^2\frac{-2x}{2\sqrt{12-x^2}}+2x\sqrt{12-x^2}\right)=\frac{3}{2}\left(\frac{8x-x^3}{\sqrt{12-x^2}}\right)$$
Now, for maximum or minimum, we have $\frac{dV}{dx}=0$ 
$$\frac{3}{2}\left(\frac{8x-x^3}{\sqrt{12-x^2}}\right)=0\implies 8x-x^3=0$$
$$x(8-x^2)=0\implies x=0, \pm 2\sqrt 2$$
But, side, $x>0$ hence, $x=2\sqrt 2$ is acceptable 
Now, it can be checked that $\frac{d^2V}{dx^2}<0$ at $x=2\sqrt 2$ i.e. the volume of prism is maximum at $x=2\sqrt 2$  
hence setting $x=2\sqrt 2$ in (1), we get the maximum volume of the inscribed equilateral triangular prism in the sphere 
$$V_{max}=\frac{1}{2}(2\sqrt 2)^2\sqrt{12-(2\sqrt 2)^2}=\color{}{8}$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\text{maximum volume of equilateral triangular prism}=8\ \text{unit}^3}}$$
