Question

Let $a+b+c=10$, $a,b,c>0$. Then the maximum value of $a^2{b}^3{c}^5$ is

• A. ${\left(15\right)}^3\times 100$
• B. ${\left(15\right)}^3\times 10$
• C. ${\left(15\right)}^3\times 1000$
• D. ${\left(60\right)}^3$

Answer 1

The correct option is A ${\left(15\right)}^3\times 100$

$a+b+c=0$

dividing $ab$ & $c$ respect to $a^2{b}^3{c}^5$

$=\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{5}+\frac{c}{5}+\frac{c}{5}+\frac{c}{5}+\frac{c}{5}=10$

We know

$AM\ge GM$

So

$\frac{[\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{c}{5}+...5times]}{10}\ge {[\frac{a}{2}.\frac{a}{2}.\frac{b}{3}....\frac{c}{5}]}^{\frac{1}{10}}\frac{10}{10}\ge {\left[\frac{a^2{b}^3{c}^5}{2^2{3}^3{5}^5}\right]}^{\frac{1}{10}}1\ge \frac{a^2{b}^3{c}^5}{2^2{3}^3{5}^5}{2}^2{3}^3{5}^5\ge a^2{b}^3{c}^5{\left(15\right)}^3{10}^2\ge a^2{b}^3{c}^5$

So maximum value $={\left(15\right)}^3\times 100$

Answer$A$