Aspire Faculty ID #16183 · Topic: NIMCET 2011 · Just now
NIMCET 2011

The minimum value of $px + qy$ when $xy=r^2$ and $p,q,x,y$ are positive numbers is

Solution

Use AM-GM or substitution $y=\dfrac{r^2}{x}$: $px + q\dfrac{r^2}{x}$ Min occurs when derivatives equal: $px = q\dfrac{r^2}{x}$ $\Rightarrow x^2 = \dfrac{qr^2}{p}$ Compute minimum: $2r\sqrt{pq}$

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