Question

If $a,b,c>0,a{b}^2{c}^3=64$ and $a+b+c=k,$ then

• A. maximum value of $k=3\sqrt{3}$
• B. at maximum value of $k,a:b:c=3:2:1$
• C. at minimum value of $k,a:b:c=1:2:3$
• D. minimum value of $k=6\times {\left(\frac{16}{27}\right)}^{1/6}$

Answer 1

Answer: (C), (D)

The correct options are
C at minimum value of $k,a:b:c=1:2:3$
D minimum value of $k=6\times {\left(\frac{16}{27}\right)}^{1/6}$

$\frac{a+2\cdot \frac{b}{2}+3\cdot \frac{c}{3}}{6}\ge \sqrt[6]{6a{\left(\frac{b}{2}\right)}^2{\left(\frac{c}{3}\right)}^3}$
$\Rightarrow k\ge 6\times \sqrt[6]{6\frac{64}{4\times 27}}=6\times {\left(\frac{16}{27}\right)}^{1/6}$
So, the minimum value of $k=6\times {\left(\frac{16}{27}\right)}^{1/6}$

At this value $a=\frac{b}{2}=\frac{c}{3}$
$\Rightarrow a:b:c=1:2:3$