Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
I can't seem to generate the equation for me to maximise. Hope someone could help please.
The answer is $\displaystyle \frac{8r^3}{3\sqrt{3}}$
Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
I can't seem to generate the equation for me to maximise. Hope someone could help please.
The answer is $\displaystyle \frac{8r^3}{3\sqrt{3}}$
The largest 'rectangular box' will be a cube.
For any rectangular box, with dimensions: L,W,H
The length of the diagonal is $\displaystyle \sqrt{ L^2 + W^2 + H^2}$
The diagonal of the cube inside the sphere is equal to the diameter of the sphere, or twice the radius.
Since the sides of the cube are equal, lets call it 's'
and the diameter of the sphere = 2r
$\displaystyle (2r)^2 = s^2 + s^2 + s^2 $
$\displaystyle (2r)^2 = 3s^2 $
$\displaystyle \sqrt{\dfrac{(2r)^2}{3}} = s = \dfrac{2r}{\sqrt{3}}$
The Volume of the cube is s^3
substituting the value above for s
Volume = $\displaystyle \left(\dfrac{2r}{\sqrt{3}}\right)^3 $
$\displaystyle \dfrac{ 2^3 r^3}{3\sqrt{3}} \, = \, \dfrac{8r^3}{3\sqrt{3}}$
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