<s>[INST] Which body has a minumum volume I would really appretiate a clear explanation.
Solve using lagrange functio.Which body, bounded by the planes $x = 0, y = 0, z = 0$ and the plane tangent to the
ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
has the minimal volume?
Solution : We need to minimiz $$V= \frac{abc}{6xyz}$$ knowing that $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$ and we will solve it using Lagrange function.
However i do not understand how do we know that $V= \frac{abc}{6xyz}$? 
And finding tangent plane $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^1} = 1$, we know that tangent has an equation $$\frac{2x_0}{a^2}(x-x_0)+\frac{2y_0}{b^2}(y-y_0)+\frac{2z_0}{c^2}(z-z_0)$$ and then how do we get that tangent plane is $$\frac{xx_0}{a^2}+\frac{yy_0}{b^2}+\frac{zz_0}{c^2}$$? and why did we put point (1,1,1) to solve the problem?
 [/INST] The body $B$ of which we want to evaluate the volume as for vertices the origin and the intersections of the plane $P$ tangent to the ellipsoid with the coordinates axes.
Knowing the equation of $P$ that you gave in your question, the $x$-axis of coordinates has for equations $y=z=0$. Plug that with the equation of $P$, you get 
$$\frac{2x}{a^2}(X-x)-\frac{2y}{b^2}y-\frac{2 z}{c^2}z=0$$
As $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$, you obtain
$$X=\frac{a ^2}{x}.$$ And similarly for the other vertices
$$Y=\frac{b ^2}{y}, \, Z=\frac{c ^2}{z}.$$
You then get the formulae
$$V =\frac{1}{3}\left[\left(\frac{a^2 b^2}{2xy}\right) \times \frac{c^2}{z}\right]=\frac{a^2 b^2 c^2}{6xyz}$$
as the volume of a tetrahedron is a third of the basis area multiplied by the height.
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