Question

Let $p,q,rϵR^{+}$ 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. $1$

Answer 1

The correct option is A $3$

$27pqr\ge (p+q+r{)}^3$

$\Rightarrow (pqr{)}^{1/3}\ge (\frac{p+q+r}{3})$

$\Rightarrow p=q=r$

Also $3p+4q+5r=12\Rightarrow p=q=r=1$

$p^3+q^4+r^5=3$

Hence, option 'A' is correct.