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(C) Given $x+y=20$. We want to maximize $f(x, y) = x^3 y$. Using the $AM$-$GM$ inequality for the terms $\frac{x}{3}, \frac{x}{3}, \frac{x}{3}, y$: $\

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$\frac{\alpha}{\beta}$

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(C) Given $x+y=20$. We want to maximize $f(x, y) = x^3 y$. Using the $AM$-$GM$ inequality for the terms $\frac{x}{3}, \frac{x}{3}, \frac{x}{3}, y$: $\frac{\frac{x}{3} + \frac{x}{3} + \frac{x}{3} + y}{4} \geq \sqrt[4]{\left(\frac{x}{3}ight)^3 y}$ $\frac{x+y}{4} \geq \sqrt[4]{\frac{x^3 y}{27}}$ $\frac{20}{4} \geq \sqrt[4]{\frac{x^3 y}{27}} \Rightarrow 5 \geq \sqrt[4]{\frac{x^3 y}{27}}$ $5^4 \geq \frac{x^3 y}{27} \Rightarrow x^3 y \leq 27 	imes 625 = 16875$. So,$k = 16875$. The maximum occurs when $\frac{x}{3} = y$. Since $x+y=20$,we have $3y+y=20$ $\Rightarrow 4y=20$ $\Rightarrow y=5=\beta$ and $x=15=\alpha$. Now,$\frac{k}{\alpha^2 \beta^2} = \frac{27 	imes 625}{15^2 	imes 5^2} = \frac{27 	imes 625}{225 	imes 25} = \frac{16875}{5625} = 3$. Since $\frac{\alpha}{\beta} = \frac{15}{5} = 3$,the correct option is $\frac{\alpha}{\beta}$.

If $x, y$ are two positive integers such that $x+y=20$ and the maximum value of $x^3 y$ is $k$ at $x=\alpha, y=\beta$,then $\frac{k}{\alpha^2 \beta^2} =$

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