Applications Of Derivatives Question 426
Question: What is the minimum value of $ px+qy $ $ (p>0,q>0) $ when $ xy=r^{2} $ -
Options:
A) $ 2r\sqrt{pq} $
B) $ 2pq\sqrt{r} $
C) $ -2r\sqrt{pq} $
D) $ 2rpq $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given that $ xy=r^{2} $
$ \Rightarrow y=\frac{r^{2}}{x} $ Let $ S=px+qy=px+\frac{qr^{2}}{x} $
$ \Rightarrow \frac{dS}{dx}=p-\frac{qr^{2}}{x^{2}} $
$ \frac{dS}{dx}=0 $ for maximum or minimum. So, $ 0=p-\frac{qr^{2}}{x^{2}} $
$ \Rightarrow x^{2}=\frac{qr^{2}}{p}\Rightarrow x=\pm \sqrt{\frac{q}{p}}.r $ Now, $ \frac{d^{2}S}{dx^{2}}=\frac{2qr^{2}}{x^{3}} $ At $ x=+\sqrt{\frac{q}{p}}.r\frac{d^{2}S}{dx^{2}}>0 $
Hence, S is minimum at $ x=\sqrt{\frac{q}{p}}.r $
$ \Rightarrow y=\frac{r^{2}}{\sqrt{\frac{q}{p}}.r}=\sqrt{\frac{p}{q}}.r $ Minimum value of $ px+qy=p.\sqrt{\frac{q}{p}}.r+q\sqrt{\frac{p}{q}}.r $
$ =\sqrt{pq}r+\sqrt{pq},r=2r\sqrt{pq} $
BETA
