If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You want to find the maximum value of this. This can be simplified as $\frac{-5x^2 + 100x}{3}$. I factored out a $-5$ to get $\frac{-5(x^2 - 20x)}{3}$. Completing the Square, I got $\frac{-5((x - 10)^2 - 100)}{3}$, or $\frac{-5(x - 10)^2 + 500}{3}$. The maximum value of this is when $x = 10$, since it makes $-5(x - 10)^2$ equal $0$ ($0$ is the greatest value because otherwise it would be negative). So my answer was $\boxed{\frac{500}{3}}$, but I'm pretty certain that isn't correct because the product of two positive integers can't be a fraction. Can someone help me out?
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EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.

Since $x$ and $y$ are positive integers, we can use AM-GM inequality. Also, notice that $3y = 5(20 -x)$, therefore $y$ is a multiple of $5$ or $xy$ is a multiple of $5$.
So, $$\frac{5x + 3y}{2} \ge \sqrt{15xy}  $$
or, $$50 \ge \sqrt{15xy}$$
By squaring both sides, we ge:$$2500 \ge 15xy$$
or $$166.67 \ge xy$$
Since $xy$ is an integer  less than $166.67$ and a multiple of $5$, so you can check that the greatest value of $xy$ is $165$.
