Question

Let a plane $3x+by+c^{2}z=12(b,c>0)$ meet the co-ordinate axes at $A,B$ and $C$ respectively such that $b^2+2c^2=48$. If minimum volume of the tetrahedron $OABC$ 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

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. $\ge$ GM.
$\frac{b^2+c^2+c^2}{3}\ge {\left(b^2\cdot c^2\cdot c^2\right)}^{\frac{1}{3}}$
${\left(b^2{c}^4\right)}^{\frac{1}{3}}\le 16$
$b^2{c}^4\le (16{)}^3\Rightarrow b{c}^2\le 64$
$\therefore {V|}_{\min .}=\frac{96}{64}=\frac{3}{2}\equiv \frac{p}{q}$
$\therefore |p-q|=1$