Question

If $a,b,c$are positive real numbers such that $a+b+c=18$, find the maximum value of $a^2{b}^3{c}^4$.

Answer 1

Find the maximum value of $a^2{b}^3{c}^4$.

Given: $a+b+c=18$

Dividing $a$ into two equal parts.

$a=\frac{a}{2}+\frac{a}{2}$

Dividing $b$ into three equal parts.

$b=\frac{b}{3}+\frac{b}{3}+\frac{b}{3}$

Dividing $c$ into four equal parts.

$c=\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}$

we know that the relation A.M$\ge$G.M

$\frac{{\displaystyle \frac{a}{2}}+{\displaystyle \frac{a}{2}}+{\displaystyle \frac{b}{3}}+{\displaystyle \frac{b}{3}}+{\displaystyle \frac{b}{3}}+{\displaystyle \frac{c}{4}}+{\displaystyle \frac{c}{4}}+{\displaystyle \frac{c}{4}}+{\displaystyle \frac{c}{4}}}{9}\ge {\left[\frac{a^2}{4}\times \frac{b^3}{3^3}\times \frac{c^4}{4^4}\right]}^{\frac{1}{9}}$

$\Rightarrow$ $\frac{a+b+c}{9}\ge {\left[\frac{a^2{b}^3{c}^4}{3^3{4}^5}\right]}^{\frac{1}{9}}$

$\Rightarrow$ $\frac{18}{9}\ge {\left[\frac{a^2{b}^3{c}^4}{3^3{4}^5}\right]}^{\frac{1}{9}}$

$\Rightarrow$ $2\ge {\left[\frac{a^2{b}^3{c}^4}{3^3{4}^5}\right]}^{\frac{1}{9}}$

$\Rightarrow$ $2^9\ge \left[\frac{a^2{b}^3{c}^4}{3^3{4}^5}\right]$

$\Rightarrow$ $2^9\times 3^3\times 4^5\ge a^2{b}^3{c}^4$

$\Rightarrow$ $2^9\times 3^3\times 2^{10}\ge a^2{b}^3{c}^4$

$\Rightarrow$ $2^{19}\times 3^3\ge a^2{b}^3{c}^4$

Hence, the maximum of $a^2{b}^3{c}^4$ is $2^{19}\times 3^3$.