volume of a triangular pyramid

Question

A point $M_0$ with positive coordinates $(x_0, y_0, z_0)$, $x_0>0$, $y_0>0$, $z_0>0$ is given. A plane is drawn through this point. Find the minimal possible volume of a triangular pyramid whose faces lie in the plane mentioned above and in the coordinate planes.

Answer 1

Let the plane equation $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ then the volume is $\frac{1}{6}abc$, so (Lagrangian multipliers, $abc\to\min$). $$\begin{cases} bc-\lambda\frac{x_0}{a^2}=0\\ ac-\lambda\frac{y_0}{b^2}=0\\ ab-\lambda\frac{y_0}{c^2}=0\\ \frac{x_0}{a}+\frac{y_0}{b}+\frac{z_0}{c}=1 \end{cases}$$ $$\begin{cases} abc-\lambda\frac{x_0}{a}=0\\ abc-\lambda\frac{y_0}{b}=0\\ abc-\lambda\frac{y_0}{c}=0\\ \frac{x_0}{a}+\frac{y_0}{b}+\frac{z_0}{c}=1 \end{cases}$$ Since $\frac{x_0}{a}=\frac{y_0}{b}=\frac{y_0}{c}$ so these three $=\frac{1}{3}$, $a=3x_0$, $b=3y_0$, $c=3z_0$, $V=\frac{27}{6}x_0y_0z_0$.