Question

Let $p,q,r\in R^{+}$and $27pqr\ge (p+q+r{)}^3$ and $3p+4q+5r=12$ then $p^3+q^4+r^5$ is equal to

• A. 3
• B. 6
• C. 2
• D. None of these

Answer 1

Answer: (A)

$27pqr\ge (p+q+r{)}^3$
$\Rightarrow (pqr{)}^{1/3}\ge \frac{p+q+r}{3}$
$\Rightarrow p=q=r$
Also, $3p+4q+5r=12$
$\Rightarrow p=q=r=1$