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