Question

If $a,b,c\in R^{+}$ such that $a+b+c=27$ , then the maximum value of $a^2{b}^3{c}^4$ is equal to

• A. $2^8\cdot 3^{10}$
• B. $2^9\cdot 3^{11}$
• C. $2^{10}\cdot 3^{12}$
• D. $2^{11}\cdot 3^{13}$

Answer 1

Answer: (C)

Consider nine numbers $\frac{a}{2},\frac{a}{2},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$
Using $AM\ge GM$ , we get,
$\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}}{9}\ge {\left({\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4\right)}^{\frac{1}{9}}$
$\Rightarrow \frac{a+b+c}{9}\ge {\left(\frac{a^2{b}^3{c}^4}{2^2{3}^3{4}^4}\right)}^{\frac{1}{9}}$
$\Rightarrow {\left(\frac{27}{9}\right)}^9\ge \frac{a^2{b}^3{c}^4}{2^2{3}^3{2}^8}$
$\Rightarrow a^2{b}^3{c}^4\le 2^{10}\cdot 3^{12}$