Question

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

• A. $2^{19}\cdot 3^3$
• B. $2^{20}\cdot 3^4$
• C. $2^{18}\cdot 3^4$
• D. $2^{19}\cdot 3^4$

Answer 1

The correct option is A $2^{19}\cdot 3^3$

Given $a+b+c=18$
$\Rightarrow (\frac{a}{2}+\frac{a}{2})+(\frac{b}{3}+\frac{b}{3}+\frac{b}{3})+(\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4})=18$

Since $A.M.\ge G.M.$
$\Rightarrow \frac{2\left(\frac{a}{2}\right)+3\left(\frac{b}{3}\right)+4\left(\frac{c}{4}\right)}{9}\ge {\left[{\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4\right]}^{1/9}$
$\Rightarrow \frac{18}{9}\ge {\left[{\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4\right]}^{1/9}$
$\Rightarrow 2^9\ge \frac{a^2{b}^3{c}^4}{2^2{3}^3{4}^4}$
$\Rightarrow a^2{b}^3{c}^4\le 3^3\cdot 2^{19}$