# How to find the minimum value of $px+qy$ when $xy=r^2$?

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?

• 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

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