I am given the following Lagrange Multipliers' problem:
Find the equation of the plane that goes through $(2,2,1)$ that creates, on the first octant along with the coordinate planes, a tetrahedron with the smallest volume
Being a Lagrange exercise, I must find the function to be minimized and the constraint.
Given that the tetrahedron's volume will be $V = \frac{xyz}{6}$ we already have the function to be minimized. I am not sure, though, how to deal with the constraint, related to the point $P=(2,2,1)$ given.
Any help is highly appreciated.
Best Answer
We can formulate the problem as a constrained optimization problem and apply Lagrange multiplier. Express the plane as $$\frac xa+\frac yb+\frac zc=1$$
It simply means that the normal direction is $\vec { n } =\left<\dfrac 1a,\dfrac 1b,\dfrac 1c\right>.$ This form of the plane is convenient for this problem since it readily shows that the intercepts with $x,y,z$ axis are $a,b,c$ respectively. Therefore the volume of the cutoff tetrahedron is $V(a,b,c)=\dfrac{abc}{6}.$ Now we need to satisfy the constraint that the plane passes a giving point. Plugging the point $(2,2,1)$ into the plane we get $\dfrac 2a+\dfrac2b+\dfrac1c=1$
So the mathematical model is a constraint optimization problem
$$\begin{cases}min_{a,b,c>0}f(a,b,c)=6V(a,b,c)=abc\\ s.t \mbox{ }g(a,b,c)=\dfrac1a+\dfrac1b+\dfrac1c-1=0\\ \end{cases}$$
By Lagrange multiplier, taking partial derivatives w.r.t. $a, b, c$ as in
$$\nabla f=\lambda\nabla g$$ We get $$bc=-2\lambda a^{-2}$$ $$ac=-2\lambda b^{-2}$$ $$ab=\lambda c^{-2}$$ Multiplying respectively $a, b, c$ on both sides of the above three equations, we get $$abc=-\lambda\frac2a=-\lambda\frac2b=-\lambda\frac1c$$
Since $a, b, c$ are positive, we have $\lambda\ne0.$ So we must have
$$a=b, a=2c$$
Plugging the above into the constraint $\dfrac2a+\dfrac2b+\dfrac1c=1$ we get $a=6,b=6,c=3.$ There for the plane that cuts off the smallest volume is $$\frac x6+\frac y6+\frac z3=1\mbox{ or } x+y+2z=1$$
And the smallest cutoff volume is $V(6,6,3)=\dfrac16\times6\times6\times3=18$