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: Since $$3y= 100-5x\implies 3\mid 20-x$$ we have $20-x =3t$ for some integer $t$. Then $x= 20-3t$ and $y= 5t$. So we have $$xy = 5t(20-3t) \leq \max \{20\cdot 8, 15\cdot 11\}$$
(Maximum value of quadratic function is at $10\over 3$ so in the set of integers it is at $3$ or at $4$.)
A: $ 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 $
A: 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 $165$.
A: $We\, multiply\, \, both\, sides\, \, by\, y.\, Then\, 5xy+3y^2=100y. \, All\, we\, have\, to\, do\, is\, maximise\, 100y-3y^2.\, Using\, calculus\, we \, obtain\, y=\frac{50}{3},\, But\, this\, is\, not\, an\, integer.\, We\, \, try\, \, 16 \, which\, is\, not\, acceptable.\, \, Then\, try\, 15 \, which\, gives\, \, x=11\, y=15\, which\, \, \, are\, acceptable.\, This\:\, \,  is\, the\, maximizer.\, Thus \, max(xy)=165.\, It\, is\, very\, easy\, to\, prove\, that\, for\, y\leq 14\,\, \,  we \, either\, get\, unacceptable\, values\, or\,\, an \, xy\, smaller\, than\, 165.Also\, \, \, \, 3y=-5x+100=5k\, and\, hence\, y\, is\, a\, multiple\, of\, 5.\, It\, is \, easy\, to\, reject\, y=20\, and\, y=25\, and\, y=30.\, That \, completes\, the\, proof!!$
A: 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
A: 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.
