<s>[INST] 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.
 [/INST] $ 5x + 3y = 100 $
Has the solution $ x = 20 - 3 t $ and $ y = 5 t $ where $ t= 1, 2, \dots $
Maximum $t $ is $\text{int}(\frac{20}{3}) = 6 $
$xy = 5 t (20 - 3 t) $
Its peak is at $\frac{1}{2} ( 0 + \frac{20}{3} ) = \frac{10}{3} $
Since this not an integer, the maximum is attained at $t = 3$ (because it closer to $\frac{10}{3}$ than $4$).  Hence, the maximum is
$ \max(xy) = 15(11) = 165 $
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