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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? ~

EDIT: I found a case where $x=11$. Then, the product is $165$. Not sure if that is the maximum, though.

  • If $x$ and $y$ are positive integers, then how can the maximum value of their product, i.e., $xy$, be $\frac{500}{3}$? – John Omielan Jun 09 '22 at 05:33
  • $500/3$ isn't an integer. – IntegrateThis Jun 09 '22 at 05:36
  • Using modular arithmetic I'm getting $xy=125$. Is modular arithmetic allowed? – Andrew Sotomayor Jun 09 '22 at 05:40
  • @AndrewSotomayor sure, you can use modular arithmetic if you want, but I'm pretty sure $125$ isn't the right answer because I found a case where $x=11$ and $y=15$, and their product is already $165$. I just need someone to help solve the question... –  Jun 09 '22 at 05:43
  • What's the minimal value of x such that $((100-5x)/3)$ is an integer? – IntegrateThis Jun 09 '22 at 05:44
  • Oh wait, what if $x=11$? Then $11-10=1$, $1^2=1$, and $(-5)(1)=-5$. $500$ minus $5$ equals $495$, and that is divisible by $3$ – you get $165$. Is that the maximum? –  Jun 09 '22 at 05:47
  • @chantria Note the solution is where $\lvert x - 10 \rvert$ is a minimum and $x$ is a positive integer solution to $5x + 3y = 100$. – John Omielan Jun 09 '22 at 05:49
  • @JohnOmielan |$x-10$| can't equal $0$, or the product would be a fraction. So the next minimum value would be when $x=9$ or when $x=11$. –  Jun 09 '22 at 05:51
  • @chantria right, I see my mistake, the correct answer is indeed 165, My solution is very similar to the one just posted by nonuser. – Andrew Sotomayor Jun 09 '22 at 05:51
  • @chantria Yes! Thus, what do you conclude from that? – John Omielan Jun 09 '22 at 05:52
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    @JohnOmielan I got the answer! Thanks for your help! –  Jun 09 '22 at 05:53
  • @chantria It's great you got the answer, and you're welcome for the help. However, one thing to be careful of though is that, in general, the solution is not necessarily the value of $x$ which minimizes $\lvert x - 10 \rvert$ and which has $xy$, i.e., $\frac{-5(x-10)^2+500}{3}$, is an integer. It works in your particular case, since for $x = 11$ we have $y = 15$, but it may not if, with a different example, we ended up with $x = 65$ and $xy = 169$ since then $y = \frac{13}{5}$, but this is not a positive integer. – John Omielan Jun 09 '22 at 06:03

6 Answers6

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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 $4$.)

nonuser
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$ 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 $

Hosam Hajeer
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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 $165$.

VVR
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$We\, multiply\, \, both\, sides\, \, by\, y.\, Then\, 5xy+3y^2=100y. \, All\, we\, have\, to\, do\, is\, maximise\, 100y-3y^2.\, Using\, calculus\, we \, obtain\, y=\frac{50}{3},\, But\, this\, is\, not\, an\, integer.\, We\, \, try\, \, 16 \, which\, is\, not\, acceptable.\, \, Then\, try\, 15 \, which\, gives\, \, x=11\, y=15\, which\, \, \, are\, acceptable.\, This\:\, \, is\, the\, maximizer.\, Thus \, max(xy)=165.\, It\, is\, very\, easy\, to\, prove\, that\, for\, y\leq 14\,\, \, we \, either\, get\, unacceptable\, values\, or\,\, an \, xy\, smaller\, than\, 165.Also\, \, \, \, 3y=-5x+100=5k\, and\, hence\, y\, is\, a\, multiple\, of\, 5.\, It\, is \, easy\, to\, reject\, y=20\, and\, y=25\, and\, y=30.\, That \, completes\, the\, proof!!$

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    Tempted to vote this down for poor use of $\LaTeX$. Would you take time to consider how it ought to be used, by inspecting how other people use it? This may be doable by using "edit" on other users' answers -- although I am assuming you have the level of authorisation to be able to do that, which you may not of course. I can't remember what you can do when you only have 44 rep. – Prime Mover Jun 09 '22 at 07:47
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    @GeorgeTsoutsinos FYI, to learn how to use the math formatting system used here, i.e., MathJax (which, as Prime Mover has pointed out, is very similar to $\LaTeX$), please see Short and helpful advice on using MathJax on the site ... and MathJax basic tutorial and quick reference. – John Omielan Jun 09 '22 at 08:06
  • Excuse me sir but I dont see any problems in my Latex, This is a mathematical site not a typing class. About my reputation score , I just joined the club and I have other occupations too!! –  Jun 09 '22 at 08:23
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    Your post would be easier to read if you only enclosed the equations and mathematical expressions between dollar signs. – N. F. Taussig Jun 09 '22 at 08:40
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    And it would be easier to write as well without all that $\backslash,$ – pH 74 Jun 09 '22 at 12:34
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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

Amit Kumar Basistha
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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.

MathematicianByMistake
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