Question

If $x+y=60,x>0,y>0$, then the maximum value of $x{y}^3$ is

• A. $(15{)}^4\frac{25}{3}$
• B. $45(15{)}^3$
• C. $\frac{(45{)}^{3}9}{5}$
• D. $\frac{(45{)}^4}{3}$

Answer 1

Answer: (D)

Given, $x+y=60$
By $AM\ge GM$
$\frac{x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}}{4}\ge {\left(\frac{x{y}^3}{27}\right)}^{1/4}$
$\Rightarrow \frac{x+y}{4}\ge {\left(\frac{x{y}^3}{27}\right)}^{\frac{1}{4}}$
$\Rightarrow \frac{60}{4}\ge {\left(\frac{x{y}^3}{27}\right)}^{\frac{1}{4}}$
$\Rightarrow \frac{x{y}^3}{27}\le (15{)}^4$
$\Rightarrow x{y}^3\le 3^3\times 3^4\times 5^4\Rightarrow x{y}^3\le \frac{(45{)}^4}{3}$
$\therefore$ Maximum value of $x{y}^3$ is $\frac{(45{)}^4}{3}$.