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The question says: "Find the minimum value of $px+qy$ when $xy=r^2$."

No information is given on $p,q,x,\text{and }y.$ However assuming the obvious I tried using this, but I am not able reduce it to the desired answer, which is $2pq\sqrt{3}$.

Any ideas?

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    The desired answer does not make sense. Where does the 3 come from? and where did the $r$ go?2011-11-30
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    No idea!; You may like to see [here](http://www.kingseducation.in/home/file.php/1/NIMCET/nimcet_2011_paper_key.pdf) (#16); the answer key is in the last page.2011-11-30
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    The answer is (1) not (2).2011-11-30
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    looking at how the sheet is ridden with mistakes, if possible, I would avoid using it to study.2011-11-30
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    I do not have advice for you, it looks like luck is needed on your part to do well for the test(assuming that the writers know the solutions to their own questions). The 'good' news is that the questions don't seem too difficult!2011-11-30
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    Are these genuine questions from a real exam? If the same people will design your exam, it may be good to have practice solving very badly worded problems.2011-11-30
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    @picakhu:I deleted my initial comment due to avoid any sort of unintended dissension.And yes there is no way to even know that have entered the correct options in the OMR machine.2011-11-30
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    @André Nicolas:Yes Andre, this is the genuine paper from a real exam, btw I don't have much idea how to practice such.2011-11-30
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    @MaX: Then practicing on the real questions is the way to go. Exam questions tend not to change fundamentally from year to year. You will need to solve too many of these questions in too little time, so you need to develop a deep familiarity with possible questions.2011-11-30

2 Answers 2

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By am-gm, we have

$px+qy \geq 2 \sqrt{pqxy} =2r \sqrt{pq}$

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    I got this same answer initially, but when the key didn't matched I thought of using something else, but aren't we suppose to use weighted AM-GM here?2011-11-30
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    You could use any sort of am-gm. You just need to check that the min can be reached. (Ross does that)2011-11-30
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    I agree; but this one is from the options so I think this is correct here.2011-11-30
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    @max, I am not sure what you mean.2011-11-30
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    I mean there is infinite number of solutions for this problem but only the option (1) matches.2011-11-30
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    I think there would be less of a dispute if the question were written as "Find the minimum value of $px + qy$ when $(px)(qy) = pqr^2$.2011-11-30
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    @Zarrax, that would make it rather redundant I think.2011-11-30
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Blindly assuming all the variables are greater than $0$ (otherwise you can send it to $-\infty$) you can write $px+qy=px+qr^2/x$. Differentiating and setting to zero gives $x=qr/p$ and plugging in gives $px+qy=2r\sqrt{pq}$. Is your $\sqrt{3}$ supposed to be $r$?