Question

Let a plane $3 x + by + c ^2 z =12( b , c >0)$ meet the co-ordinate axes at $A , B$ and $C$ respectively such that $b^2+2 c^2=48$. If minimum volume of the tetrahedron $O A B C$ where ' $O$ ' is the origin is equal to $\frac{p}{q}$, where $p, q \in N$, then find the least value of $|p-q|$.

Answer 1

Volume of the tetrahedron $OABC $ is
$V=\frac{1}{6} \cdot 4 \cdot \frac{12}{b} \cdot \frac{12}{c^2}=\frac{96}{b^2}$
Applying, A.M. $\geq$ GM.
$ \frac{ b ^2+ c ^2+ c ^2}{3} \geq\left( b ^2 \cdot c ^2 \cdot c ^2\right)^{\frac{1}{3}}$
$ \left( b ^2 c ^4\right)^{\frac{1}{3}} \leq 16 $
$ b ^2 c ^4 \leq(16)^3 \Rightarrow bc ^2 \leq 64$
$\therefore \left. V \right|_{\min .}=\frac{96}{64}=\frac{3}{2} \equiv \frac{ p }{ q } $
$\therefore | p - q |=1$