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I'm studying for my calc II exam, and I recently discovered how one can solve max/min problems with constrains using AM-GM inequalities instead of lagrange multipliers. Seeing as this is way more elegant, I'd like to learn more about it.

Could someone help me find the maximum and minimum value of $3+2xy$ subject to the constraint $x^2+y^2=1$. I've already solved it using lagrange multipliers, but I'd like to see how it can be solved using AM-GM aswell.

(I've only looked at the very basics of AM-GM so far, but I couldn't find much online about how to utilize the inequalities when you have a constraint. Is this worth learning before my exam?)

Thank you!

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  • $\begingroup$ Are $x,y$ assumed to be positive? $\endgroup$ May 6, 2018 at 16:38
  • $\begingroup$ Well no, the task is taken from a previous exam, lagrange was used in the solution but I thought you can use AM-GM. The question was simply to find the maximum and minimum value $3+2xy$ on the circle $x^2+y^2=1$ $\endgroup$
    – novo
    May 6, 2018 at 16:43

2 Answers 2

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Alternative, polar coordinates.

$x=r \cos t$, $y=r\sin t$, $0\le t \lt 2π.$

Note: $r =1.$

$F(r,t):= 3+2\sin t \cos t =$

$3+ \sin 2t.$

Minimum, maximum?

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  • $\begingroup$ Not quite AM-GM, but this one was quite elegant as well! Thank you for the alternative solution (without using lagrange) $\endgroup$
    – novo
    May 6, 2018 at 20:28
  • $\begingroup$ novo.Welcome.Fairly simple in polar coordinates. $\endgroup$ May 6, 2018 at 21:46
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From $$x^2+y^2=1$$ we get $$(x+y)^2=1+2xy\geq 0$$ For the Maximum prove that $$3+2xy\le 4$$

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  • $\begingroup$ I tried setting $\frac{1+2xy}{2} \geq \sqrt{2xy}$ does $\sqrt{2xy}$ simplify to 1? $\endgroup$
    – novo
    May 6, 2018 at 16:53
  • $\begingroup$ This is only correct, when $x,y$ are assumed to be positive! $\endgroup$ May 6, 2018 at 16:54
  • $\begingroup$ Ok, so if the task does not specifiy that $x, y \geq 0$ than I should stick to lagrange multipliers? If we do assume $x, y \geq 0$ can we also find the minimum? $\endgroup$
    – novo
    May 6, 2018 at 16:57
  • $\begingroup$ Yes, then you will get $$3+2xy\geq 1$$ $\endgroup$ May 6, 2018 at 17:02
  • $\begingroup$ and since $$x^2+y^2\geq 2xy$$ we get $$3+2xy\le 4$$ $\endgroup$ May 6, 2018 at 17:05

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