With the given point $M(a,b,c)$ in $\mathbb R^3$, find the tetrahedron with the smallest volume that is formed with a plane that contains $M$ and who's points are the intersections of that plane with the coordinate axises.
Lets say the plane is $P:Ax+By+Cz+D=0.$ (are the intersections of this plane with the axises $x,y,z$: $A,B,C$ in order?) Either way, since $M \in P$ $Aa+Bb+Cc+D=0.$ If the intersections are $m,n,p$ with $x,y,z$ then the volume of the tetrahedron should be $$\frac{1}{6}|(m-n)||(n-p)|(m-p)|$$
I don't know how to set up the Lagrange equation for this problem. Can anyone help out? Exam tomorrow.
We can see that $Aa+Bb+Cc+D=0$, so the plane $P:Ax+By+Cz+D=0$ can be written: $$P:A(x-a)+B(y-b)+C(z-c)=0$$ Consider the intersection of $P$ and z-axis. All points on z-axis has the form $(0,0,p)$, then $$A(0-a)+B(0-b)+C(p-c)=0\iff p=\frac{Aa+Bb+Cc}{C}$$ Similarly, two another intersection has coordinate $(\frac{Aa+Bb+Cc}{A},0,0)$ and $(0,\frac{Aa+Bb+Cc}{B},0)$.
The volume of the tetrahedron is $$(Aa+Bb+Cc)^3\over6ABC$$ WLOG assume $ABC=1$ (otherwise we can scale A, B and C by a constant factor). So we have to minimize $Aa+Bb+Cc$ with the respect $ABC=1$.
That can be done as follow: $$\frac{Aa+Bb+Cc}{3} \geq AaBbCc^{1\over3} = abc^{1\over3}$$ (AM-GM inequality)
The equality hold at $Aa = Bb = Cc$.