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.

A not very strict approach using a rather standard geometric reuslt is as follows.
$5x+3y=100$ depicts a line on the plane, specifically the $y=-\frac{5}{3}x-\frac{100}{3},$ which intersects the $y-$axis on $A=(0,33,333)$ and the $x-$axis on $B=(20,0)$.
Thus the right triangle $\triangle AOB$ is formed, where $O=(0,0)$.
Now, the desired greatest possible value of $xy$ is equivalent to the area of a rectangle inside $\triangle AOB$.
But we know that the maximum area of a rectangle inscribed in a triangle is  1/2  the area of the triangle.
By a direct computation, $E_{\triangle AOB}=333,33$. Moreover $\triangle AOB$'s perpendicular sides are of a similar length. Thus the maximum of $xy$ is expected to be near the value of $\frac{1}{2}E_{\triangle AOB}=166,67$ with $x,y$ also of approximately similar lengths.
By experimenting with pairs of $x,y$ such that $xy\approx167$, we readily observe that $(11,15)$ and $(15,11)$ are the optimal. But the second pair doesn't belong to $y=-\frac{5}{3}x-\frac{100}{3}$ and thus $(x,y)=(11,15)$ is the answer.
