Question

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

• A. $2^{18}{3}^2.4$
• B. $2^{17}.3^2.4^2$
• C. $2^{19}{3}^3.$
• D. $2^{18}{3}^3.$

Answer 1

The correct option is C $2^{19}{3}^3.$

Given $2.\frac{a}{2}+3.\frac{b}{3}+4.\frac{c}{4}=18$

Now

$AM\ge GM$

$\frac{2.\frac{a}{2}+3.\frac{b}{3}+4.\frac{c}{4}}{9}\ge {\left({\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4\right)}^{\frac{1}{9}}$

${\left(\frac{18}{9}\right)}^9\ge {\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4$

$2^9{.2}^2{.3}^3{.4}^4\ge a^2{b}^3{c}^4$

$\therefore a^2{b}^3{c}^4\le 2^{19}{.3}^3$