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.

Sometimes while finding maxima for positive integers you will have to implement some other techniques along with the traditional ones.In your problem you get $xy=\frac{-5(x-10)^2+500}{3}$.Note that $500\equiv 2$(mod 3) and $-5\equiv 1$(mod 3).So we must have $(x-10)^2\equiv 1$(mod 3).So x-10 can't be a multiple of 3.So for minimum value x=11 and thus xy=165
