Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The desired answer does not make sense. Where does the 3 come from? and where did the $r$ go? – picakhu Nov 30 '11 at 16:08
No idea!; You may like to see here (#16); the answer key is in the last page. – Quixotic Nov 30 '11 at 16:12
The answer is (1) not (2). – picakhu Nov 30 '11 at 16:14
looking at how the sheet is ridden with mistakes, if possible, I would avoid using it to study. – picakhu Nov 30 '11 at 16:17
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! – picakhu Nov 30 '11 at 16:23
up vote 5 down vote accepted

By am-gm, we have

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

share|cite|improve this answer
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? – Quixotic Nov 30 '11 at 16:14
You could use any sort of am-gm. You just need to check that the min can be reached. (Ross does that) – picakhu Nov 30 '11 at 16:15
I agree; but this one is from the options so I think this is correct here. – Quixotic Nov 30 '11 at 16:17
@max, I am not sure what you mean. – picakhu Nov 30 '11 at 16:19
I mean there is infinite number of solutions for this problem but only the option (1) matches. – Quixotic Nov 30 '11 at 16:21

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 $r$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.