Answer
The plane that minimizes $V$ if the plane is constrained to pass through the point $P = \left( {\alpha ,\beta ,\gamma } \right)$ is
$\frac{x}{{3\alpha }} + \frac{y}{{3\beta }} + \frac{z}{{3\gamma }} = 1$
Work Step by Step
With the same set-up as in Exercise 35, we are given a tetrahedron formed by the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ ($a,b,c > 0$) with the positive coordinate planes ($x,y,z > 0$).
Our task is to find $a$, $b$, and $c$ such that the volume of the tetrahedron $V = \frac{1}{6}abc$ is minimum among all planes passing through the point $P = \left( {\alpha ,\beta ,\gamma } \right)$ with $\alpha ,\beta ,\gamma > 0$.
Since the plane must pass through $P = \left( {\alpha ,\beta ,\gamma } \right)$, the equation of the plane becomes
$\frac{\alpha }{a} + \frac{\beta }{b} + \frac{\gamma }{c} = 1$
Thus, our problem can be stated: minimize $V\left( {a,b,c} \right) = \frac{1}{6}abc$ subject to the constraint $g\left( {a,b,c} \right) = \frac{\alpha }{a} + \frac{\beta }{b} + \frac{\gamma }{c} - 1 = 0$.
Step 1. Write out the Lagrange equations
Using Theorem 1, the Lagrange condition $\nabla V = \lambda \nabla g$ yields
$\left( {\frac{{bc}}{6},\frac{{ac}}{6},\frac{{ab}}{6}} \right) = \lambda \left( { - \frac{\alpha }{{{a^2}}}, - \frac{\beta }{{{b^2}}}, - \frac{\gamma }{{{c^2}}}} \right)$
So, the Lagrange equations are
(1) ${\ \ \ \ }$ $\frac{{bc}}{6} = - \frac{{\alpha \lambda }}{{{a^2}}}$, ${\ \ }$ $\frac{{ac}}{6} = - \frac{{\beta \lambda }}{{{b^2}}}$, ${\ \ }$ $\frac{{ab}}{6} = - \frac{{\gamma \lambda }}{{{c^2}}}$
Step 2. Solve for $\lambda$ in terms of $x$ and $y$
Since $a,b,c > 0$, so
$\lambda = - \frac{{{a^2}bc}}{{6\alpha }} = - \frac{{a{b^2}c}}{{6\beta }} = - \frac{{ab{c^2}}}{{6\gamma }}$
$\frac{{{a^2}bc}}{\alpha } = \frac{{a{b^2}c}}{\beta } = \frac{{ab{c^2}}}{\gamma }$
Step 3. Solve for $x$ and $y$ using the constraint
From Step 2 we obtain $\frac{{{a^2}bc}}{\alpha } = \frac{{a{b^2}c}}{\beta }$ and $\frac{{a{b^2}c}}{\beta } = \frac{{ab{c^2}}}{\gamma }$.
From the first and the second equation we obtain $\frac{a}{\alpha } = \frac{b}{\beta }$ and $\frac{b}{\beta } = \frac{c}{\gamma }$, respectively. Thus,
$\frac{a}{\alpha } = \frac{b}{\beta } = \frac{c}{\gamma }$
$b = a\frac{\beta }{\alpha }$, ${\ \ \ }$ $c = a\frac{\gamma }{\alpha }$
Substituting these in the constraint gives
$\frac{\alpha }{a} + \frac{\beta }{b} + \frac{\gamma }{c} - 1 = 0$
$\frac{\alpha }{a} + \frac{\alpha }{a} + \frac{\alpha }{a} = 1$
So, $a = 3\alpha $.
Since $b = a\frac{\beta }{\alpha }$ and $c = a\frac{\gamma }{\alpha }$, the critical point is $\left( {a,b,c} \right) = \left( {3\alpha ,3\beta ,3\gamma } \right)$.
Step 4. Calculate the critical values
So, the extreme value of $V$ at $\left( {a,b,c} \right) = \left( {3\alpha ,3\beta ,3\gamma } \right)$ is $V = \frac{{9\alpha \beta \gamma }}{2}$.
Since the volume $V = \frac{1}{6}abc$ depends on $a$, $b$ and $c$, we can increase any edge of the tetrahedron as large as possible but at the same time adjust the other edges such that the plane still pass through the point $P = \left( {\alpha ,\beta ,\gamma } \right)$. Hence, the volume can be as large as possible. Thus, we conclude that the extreme value $\frac{{9\alpha \beta \gamma }}{2}$ is the minimum value of $V$ subject to the constraint.
Since $\left( {a,b,c} \right) = \left( {3\alpha ,3\beta ,3\gamma } \right)$, the plane that minimizes $V$ if the plane is constrained to pass through the point $P = \left( {\alpha ,\beta ,\gamma } \right)$ is
$\frac{x}{{3\alpha }} + \frac{y}{{3\beta }} + \frac{z}{{3\gamma }} = 1$
Notice that there is no maximum value of $V$ since there are edges on a plane that are arbitrarily far from the origin.
