Question

If $a+b+c=18$ and $a,b,c\in R^{+}$ then the greatest value of $a^2{b}^3{c}^4$, is

• A. $2^{19}\times 3^2$
• B. $2^{19}\times 3^3$
• C. $2^{18}\times 3^3$
• D. None of the above.

Answer 1

The correct option is B $2^{19}\times 3^3$

given $a+b+c=18$

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

$⟹\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}}{9}\ge \sqrt[9]{9\frac{a}{2}\ast \frac{a}{2}\ast \frac{b}{3}\ast \frac{b}{3}\ast \frac{b}{3}\ast \frac{c}{4}\ast \frac{c}{4}\ast \frac{c}{4}\ast \frac{c}{4}}$

$⟹\frac{\frac{2a}{2}+\frac{3b}{3}+\frac{4c}{4}}{9}\ge \sqrt[9]{9\frac{a^2}{4}\ast \frac{b^3}{3^3}\ast \frac{c^4}{4^4}}$

$⟹\frac{a+b+c}{9}\ge \sqrt[9]{9\frac{a^2{b}^3{c}^4}{4^5\ast 3^3}}$

$⟹\frac{18}{9}\ge \sqrt[9]{9\frac{a^2{b}^3{c}^4}{(2^2{)}^5\ast 3^3}}$

$⟹2\ge \sqrt[9]{9\frac{a^2{b}^3{c}^4}{2^{10}\ast 3^3}}$

$⟹\sqrt[9]{9\frac{a^2{b}^3{c}^4}{2^{10}\ast 3^3}}\le 2$

taking power 9 on both sides

$⟹\frac{a^2{b}^3{c}^4}{2^{10}\ast 3^3}\le 2^9$

$⟹a^2{b}^3{c}^4\le 2^9\ast 2^{10}\ast 3^3$

$⟹a^2{b}^3{c}^4\le 2^{19}\ast 3^3$

Therefore the greatest value of $a^2{b}^3{c}^4$ is $2^{19}\ast 3^3$