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Let $K \in R^3$ the ellipsoid given by the equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $ with $a,b,c > 0$ , let $(x,y,z) \in K$ on the first octant, consider the parallelepiped of vertices $(\pm x,\pm y,\pm z)$ inscribed on $K$ with volume $V = 8xyz$.

How can I find the maximum possible value of $V$?

I stuck with this hard problem for me i tried to find the explicit equation and then get the maximum values : Let $P=(x,y,z)$ be a point on the ellipsoid with $x,y,z\gt 0$.Then i took the eight different points with $P_i (\pm x,\pm y,\pm z)$ the vertices of a parallelepiped with the side length $2x , 2y$ and $2z$.

Then, the volume parallelepiped is $V = 2x\cdot 2y\cdot 2z = 8 xyz$ and i remembered that $V$ is maximum if and only if $V^2$ is maximum .

Some help please.

Martin Sleziak
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Rosa Maria Gtz.
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1 Answers1

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Because

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$

We have

$$\left(\frac{x^2}{a^2}\frac{y^2}{b^2}\frac{z^2}{c^2}\right)^{1/3} \le \frac{1}{3}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = \frac{1}{3}$$

Thus $$V=8|xyz| \le \frac{8abc}{\sqrt{27}}$$

mike
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  • why you get $$\left(\frac{x^2}{a^2}\frac{y^2}{b^2}\frac{z^2}{c^2}\right)^{1/3} \le \frac{1}{3}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) $$ ? – Rosa Maria Gtz. Jun 23 '14 at 04:30
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    geometric mean $\le$ arithmetic mean, like $(a+b)/2 \ge \sqrt{ab}$ – mike Jun 23 '14 at 04:43