Question

In a right angle triangle $PQR,$ right-angled at $P.$ Let $PQ=4,PR=3$ and $QR=5$. There is a point $O$ inside the triangle from which perpendicular are drawn on the sides $PQ,QR,RP$ with lengths $x,y,z$ respectively. The maximum integral value of $xyz$ is

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

The correct option is A $1$


The area of $△PQR$
$\frac{1}{2}\times 4\times 3=\frac{1}{2}\times 4\times x+\frac{1}{2}\times 3\times z+\frac{1}{2}\times 5\times y\Rightarrow 4x+3z+5y=12$
Using $A.M\ge G.M$
$\frac{4x+3z+5y}{3}\ge (4x\times 3z\times 5y{)}^{1/3}\Rightarrow 4^3\ge 60xyz\Rightarrow xyz\le \frac{4^3}{60}\Rightarrow xyz\le \frac{16}{15}$

Hence the maximum integral value of $xyz$ will be $1$