Question

A regular hexagonal prism has the surface area $S$ .The largest volume of the prism is

• A. $\frac{4\sqrt{S^3}}{3\sqrt[4]{43^3}}$
• B. $\frac{2\sqrt{S^3}}{3\sqrt[4]{43^3}}$
• C. $\frac{\sqrt{S^3}}{6\sqrt[4]{43^3}}$
• D. $\frac{4\sqrt{S^3}}{3\sqrt[5]{53^4}}$

Answer 1

The correct option is C $\frac{\sqrt{S^3}}{6\sqrt[4]{43^3}}$

Let a be the side of the base and $H$ be the height of the prism. The area of the base is given by

$A_B=\frac{a^2\sqrt{3}}{4}.6=\frac{3\sqrt{3}{a}^2}{2}$
Then the surface area of the prism is expressed by the formula
$S=2A_B+A_L=2\cdot \frac{3\sqrt{3}{a}^2}{2}+6aH$
$=3\sqrt{3}{a}^2+6aH.$
$\Rightarrow 6aH=S-3\sqrt{3}{a}^2,$
$\Rightarrow H=\frac{S-3\sqrt{3}{a}^2}{6a}=\frac{S}{6a}-\frac{\sqrt{3}a}{2}$
The volume of the prism


$=\frac{3\sqrt{3}{a}^2}{2}(\frac{S}{6a}-\frac{\sqrt{3}a}{2})$
$=\frac{\sqrt{3}aS-9a^3}{4}=V\left(a\right)$
Differentiating w.r.t. $a$
$V^{\prime }\left(a\right)={\left(\frac{\sqrt{3}aS-9a^3}{4}\right)}^{\prime }=\frac{\sqrt{3}S-27a^2}{4};$
$V^{\prime }\left(a\right)=0,\Rightarrow \frac{\sqrt{3}S-27a^2}{4}=0,\Rightarrow \sqrt{3}S-27a^2=0$
$\Rightarrow a^2=\frac{\sqrt{3}S}{27},$
$\Rightarrow a=\frac{\sqrt[4]{43}\sqrt{S}}{3\sqrt{3}}=\frac{\sqrt{S}}{3\sqrt[4]{43}}$
The second derivative is
$V"\left(a\right)=-\frac{54a}{4}=-\frac{27a}{2}$
It is negative for $a>0,$ so the point a= $\frac{\sqrt{S}}{3\sqrt[4]{43}}$ is a point of local maximum by the second Derivative Test.
Now we can calculate the maximum volume of the prism:
$V_{max}=V(a=\frac{\sqrt{S}}{3\sqrt[4]{43}})$
$=\frac{\sqrt{3}\cdot \frac{\sqrt{S}}{3\sqrt[4]{43}}\cdot S-9{\left(\frac{\sqrt{S}}{3\sqrt[4]{43}}\right)}^3}{4}$
$=\frac{3^{\frac{1}{2}}\cdot S^{\frac{3}{2}}}{12\cdot 3^{\frac{1}{4}}}-\frac{3^2\cdot S^{\frac{3}{2}}}{4\cdot 3^3{.3}^{\frac{3}{4}}}$
$=\frac{S^{\frac{3}{2}}}{4\cdot 3^{\frac{3}{4}}}-\frac{S^{\frac{3}{2}}}{4\cdot 3^{\frac{7}{4}}}=\frac{S^{\frac{3}{2}}}{4\cdot 3^{\frac{3}{4}}}(1-\frac{1}{3})$
$=\frac{1}{6}\cdot \frac{S^{\frac{3}{2}}}{3^{\frac{3}{4}}}=\frac{\sqrt{S^3}}{6\sqrt[4]{43^3}}$