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If $x+y=60, x>0, y>0$, then the maximum value of $x y^3$ is
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Verified Answer
The correct answer is:
$\frac{(45)^4}{3}$
Given, $x+y=60$
By $\mathrm{AM} \geq \mathrm{GM}$
$$
\begin{array}{ll}
& \frac{x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}}{4} \geq\left(\frac{x y^3}{27}\right)^{1 / 4} \\
\Rightarrow \quad & \frac{x+y}{4} \geq\left(\frac{x y^3}{27}\right)^{\frac{1}{4}} \\
\Rightarrow \quad & \frac{60}{4} \geq\left(\frac{x y^3}{27}\right)^{\frac{1}{4}} \\
\Rightarrow \quad & \frac{x y^3}{27} \leq(15)^4 \\
\Rightarrow \quad & x y^3 \leq 3^3 \times 3^4 \times 5^4 \Rightarrow x y^3 \leq \frac{(45)^4}{3}
\end{array}
$$
$\therefore$ Maximum value of $x y^3$ is $\frac{(45)^4}{3}$.
By $\mathrm{AM} \geq \mathrm{GM}$
$$
\begin{array}{ll}
& \frac{x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}}{4} \geq\left(\frac{x y^3}{27}\right)^{1 / 4} \\
\Rightarrow \quad & \frac{x+y}{4} \geq\left(\frac{x y^3}{27}\right)^{\frac{1}{4}} \\
\Rightarrow \quad & \frac{60}{4} \geq\left(\frac{x y^3}{27}\right)^{\frac{1}{4}} \\
\Rightarrow \quad & \frac{x y^3}{27} \leq(15)^4 \\
\Rightarrow \quad & x y^3 \leq 3^3 \times 3^4 \times 5^4 \Rightarrow x y^3 \leq \frac{(45)^4}{3}
\end{array}
$$
$\therefore$ Maximum value of $x y^3$ is $\frac{(45)^4}{3}$.
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