Question

A rectangular box with a square base is inscribed in a hemisphere of radius $R$. The maximum volume of the box is

• A. $\frac{8R^3}{3\sqrt{3}}$
• B. $\frac{4R^3}{3\sqrt{3}}$
• C. $\frac{8R^2}{3\sqrt{5}}$
• D. $\frac{\sqrt{3}{R}^3}{8}$

Answer 1

The correct option is B $\frac{4R^3}{3\sqrt{3}}$


The base of the rectangular box lies in the plane that contains the base of the hemisphere. Using the Pythagoras theorem,we can write relationship:
${(\sqrt{2}\cdot \frac{x}{2})}^2+y^2=R^2,\Rightarrow \frac{x^2}{2}+y^2=R^2$
Hence
$y=\sqrt{R^2-\frac{x^2}{2}}$
The volume of the inscribed box is given by
$V=x^{2}y=x^2\sqrt{R^2-\frac{x^2}{2}}$
$=\frac{x^2}{\sqrt{2}}\sqrt{2R^2-x^2}=V\left(x\right)$
Let $D\left(x\right)=2V^2\left(x\right)=2R^2{x}^4-x^6$
The derivative of the function $D\left(x\right)$ is written in the form
$D^{\prime }\left(x\right)=8R^2{x}^3-6x^5$
Using the First Derivative Test, we find that the function $V\left(x\right)$ has a maximum at $x=\frac{2R}{\sqrt{3}}$
Now, $\frac{d^2}{d{x}^2}\left(D\right(x\left)\right)=24R^2{x}^2-30x^4$
$\Rightarrow D\left(\frac{2R}{\sqrt{3}}\right)<0$
$\Rightarrow$ For maxima $x=\frac{2R}{\sqrt{3}}$
Calculating the value of $y$:
$y=\sqrt{R^2-\frac{x^2}{2}}=\sqrt{R^2-\frac{{\left(\frac{2R}{\sqrt{3}}\right)}^2}{2}}$
$=\sqrt{R^2-\frac{2R^2}{3}}=\sqrt{\frac{R^2}{3}}=\frac{R}{\sqrt{3}}$
Then the maximum volume of the box is equal to
$V_{max}=x^{2}y={\left(\frac{2R}{\sqrt{3}}\right)}^2\cdot \frac{R}{\sqrt{3}}$
$=\frac{4R^3}{3\sqrt{3}}$