Question

The maximum value of $a^2{b}^3{c}^4$ subject to $a+b+c=18$ is-

• A. $4^2{6}^3{8}^4$
• B. $4^3{6}^2{8}^4$
• C. $4^4{6}^2{8}^3$
• D. $4^2{6}^4{8}^3$

Answer 1

Answer: (A)

$4^2{6}^3{8}^4$

Given that, $a+b+c=18$
$\Rightarrow \frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}=18$
Since, $A.M\ge G.M$
Therefore, $\frac{1}{9}(\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4})\ge {\left(\frac{a^2{b}^3{c}^4}{2^2{3}^3{4}^4}\right)}^{\frac{1}{9}}$
$\Rightarrow \frac{18}{9}\ge {\left(\frac{a^2{b}^3{c}^4}{2^2{3}^3{4}^4}\right)}^{\frac{1}{9}}$
$\Rightarrow a^2{b}^3{c}^4\le 4^2{6}^3{8}^4$
Ans: A