Question

a, b, c are three positive numbers and $ab{c}^2$ has greatest value $\frac{1}{64}$. Then

• A. $a=b=\frac{1}{2},c=\frac{1}{4}$
• B. $a=b=\frac{1}{4},c=\frac{1}{2}$
• C. $a=b=c=\frac{1}{3}$
• D. none of these

Answer 1

Answer: (B)

$a=b=\frac{1}{4},c=\frac{1}{2}$

Using A.M $⪈$ G.M
$\Rightarrow$ $\frac{a+b+c^2}{3}$ $⪈$ $(ab{c}^2{)}^{1/3}$
given maximum value of $ab{c}^2$ is 1/64
In above equation equality only holds if all the terms are equal $\Rightarrow$ $a$ $=$ $b$ $=$ $c^2$
so, $a$ $=$ $b$ $=$ $c^2$$=$ $\frac{1}{4}$
$\Rightarrow$ $a$ $=$ $b$ $=$ $\frac{1}{4}$ and $c$ $=$ $\frac{1}{2}$
Hence, option 'B' is correct.