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I need to find the dimensions of the box with maximum volume (with faces parallel to the coordinate planes) that can be inscribed in ellipsoid

$$\frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1$$

A hint given was:

If vertex of box in first octants is (x,y,z) then volume is 8xyz.

So I first find the partial derivatives

$V_x = 8yz$

$V_y = 8xz$

$V_z = 8xy$

Now the partial derivatives for constraint:

$g_x = \lambda\frac{x}{2}$

$g_y = \lambda\frac{2y}{9}$

$g_z = \lambda\frac{z}{8}$

Then $f_\alpha = \lambda g_\alpha$ so:

$8yz = \lambda\frac{x}{2} \rightarrow \lambda = \frac{16yz}{x}$

$8xz = \lambda\frac{2y}{9} \rightarrow \lambda = \frac{36xz}{y}$

$8xy = \lambda\frac{z}{8} \rightarrow \lambda = \frac{64xy}{z}$

$\frac{16yz}{x} = \frac{36xz}{y} \rightarrow y = \frac{3x}{2}$

$\frac{36xz}{y} = \frac{64xy}{z} \rightarrow y = \frac{3z}{4}$

$\frac{3x}{2} = \frac{3z}{4} \rightarrow x = \frac{z}{2}$

Now how do I continue? I seems to be missing something?

3 Answers 3

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You can cheat here, because the property of being a maximum-volume axis-parallel box is preserved by "stretching" transformations of the form $(x,y,z) \mapsto (ax,by,cz)$ (because such a transformation preserves the property of being axis-parallel, and multiplies all volumes by the constant $abc$).

So start with a sphere of radius 1: then the biggest box has corners

$$\left(\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3},\pm\sqrt \frac{1}{3}\right)$$

and a volume of $\left(2\sqrt \frac{1}{3}\right)^3 = \frac{8}{3}\sqrt \frac{1}{3}$ (see the comments for more about this). Now stretch it by a factor of $2$ along the $x$-axis, $3$ along the $y$-axis, and $4$ along the $z$-axis. The corners go to

$$\left(\pm2\sqrt \frac{1}{3},\pm3\sqrt \frac{1}{3},\pm4\sqrt \frac{1}{3}\right)$$

and the volume is $64\sqrt \frac{1}{3}$.

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    How do you prove that the biggest box inside the sphere is the cube? I don't think you can take this step for granted.2012-04-08
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    @David: I spent 20 minutes adding a proof to my post, and then the math rendering software crashed. So I'll just say: Project onto the $(x,y)$-plane, and then you can use Euclidean geometry to prove that the $x$- and $y$-coordinates are equal (a square has the maximal area of all rectangles inside a circle). Similarly for the $x$- and $z$-coordinates.2012-04-08
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    So you've shown that "a square has maximal area of rectangles in a circle" implies that "a cube has maximal volume of boxes inside a sphere". And quite elegantly too. However, I still think you're missing a step. You need to apply either the AM/GM or QM/GM inequality to get that initial result. André Nicolas' answer is a nice elegant way of doing this.2012-04-08
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    @David: It's easy to see that a cyclic quadrilateral of maximal area must have all its sides equal: if $AB \ne BC$, then the triangle $ABC$ can be made larger by positioning $B$ at a point on the circle further from the chord $AC$.2012-04-09
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    ... and if you add that to your answer, I shall upvote it.2012-04-09
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Very recently, you asked for advice on a proof of the AM/GM inequality in three variables. This inequality says that for $u$, $v$, and $w$ all $\ge 0$, we have $$\frac{u+v+w}{3}\ge \sqrt[3]{uvw},$$ with equality iff $u=v=w$. The result you want is a direct consequence of three variable AM/GM.

Let $u=\frac{x^2}{4}$, $v=\frac{y^2}{9}$, and $w=\frac{z^2}{16}$. We are told that $u+v+w=1$. It follows from AM/GM that $$\frac{1}{3}\ge \sqrt[3]{uvw}=\sqrt[3]{\frac{x^2y^2z^2}{4\cdot 9\cdot 16}}\tag{$\ast$}$$ with equality if $\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}$. From the inequality $(\ast)$, you can reach a conclusion about the maximum value of $xyz$ for $x$, $y$, $z$ non-negative. Eight times this maximum value gives the maximum volume, as per the hint.

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    Delightful answer!2014-02-11
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It seem that you've got far enough to express $y$ and $z$ in terms of $x$, which is the hard part. Now just substitute those two expressions back into the equation of the ellipsoid, and you'll have a simple equation for $x$.