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Show that the volume of the largest rectangular parallelepiped that can be inscribed in the ellipsoid $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ is $\dfrac{8abc}{3\sqrt3}$.

I proceeded by assuming that the volume is $xyz$ and used a Lagrange multiplier to start with $$xyz+\lambda \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1\right)$$ I proceeded further to arrive at $\frac{abc}{3\sqrt3}$. Somehow I seemed to be have missed $8$. Can someone please tell me where I did go wrong?

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Maybe you need to understand the following :

Let $P=(x,y,z)$ be a point on the ellipsoid with $x,y,z\gt 0$. Take the eight different points with $$P_i (\pm x,\pm y,\pm z)$$ These points are the vertices of a parallelepiped with the side length $2x , 2y$ and $2z$. Then, the volume parallelepiped is: $$V = 2x\cdot 2y\cdot 2z = 8\cdot x\cdot y\cdot z.$$

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Let $$h(x) = \text{maximum value} =8xyz +f(x^2/a^2 +y^2/b^2 +z^2/C^2+1)\tag{1}$$ where f =langragian multiplier.

Then differintiate partialy with respect to x $$8yz+2fx/a^2 =0\tag{2}$$ $$1/a^2 =(-8yz/2fx) $$ similarily $$1/b^2 =(-8zx/2fy)$$ $$ 1/c^2=(-8xy/2fz)$$ $$x^2/a^2 +y^2/b^2+z^2/c^2=1$$ $$\Rightarrow f=-12xyz$$ put f value in eqn(2) $$x=a/\sqrt 3$$ similarily $$y=b/\sqrt 3$$ $$ z=c/\sqrt 3$$ $\Rightarrow$ the largest volume of parallelopiped inscribed in ellipsoid $$=8xyz=8(a/\sqrt 3)(b/\sqrt 3)(c/\sqrt3)$$

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  • $\begingroup$ Check out the following link to help format your answer. As is, it is very difficult to read making it of little value. meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Vincent Aug 18 '14 at 20:14
  • $\begingroup$ is there any method to check weather x,y,z yield maximum volume. or we will have to find different ordered pair $\endgroup$ – Naruto_007 Aug 5 '17 at 10:58

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