Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Find the minimum value of (ax + by), where       <div style="float:left; width:48%

Find the minimum value of (ax + by), where xy = c^{2} 

 

 
 
 
 
 

Answers (1)

\\ \text { Given: } y=\frac{c^{2}}{x} \\ $Let P $ =a x+b y \\ P=a x+\frac{b c^{2}}{x} \\ $For critical point value $\frac{d P}{d x}= 0 \\ \frac{dP}{dx} = a-\frac{b c^{2}}{x^{2}}=0 \\ x^{2}=\frac{b c^{2}}{a} \Rightarrow x=\sqrt{\frac{b}{a}} \cdot c \\ $For minima $ \\ \frac{d^2P}{dx^2} = \frac{2bc^2}{x^3} \\ \\ \left.\frac{d^{2} P}{d x^{2}}\right|_{x=\sqrt{\frac{b}{a}}.c}=\frac{2 b c^{2}}{(\sqrt{\frac{b}{a}}.c)^{3}} \\=\frac{2 b }{c}\left[\sqrt{\frac{a}{b}} \right]^{3}>0

\\\text { At } x=\sqrt{\frac{b}{a}} \cdot c $ P has minimum value

\\$ minimum value of P = $ a \sqrt{\frac{b}{a}} \cdot c+ \frac{bc^{2}}{c} \sqrt{\frac{a}{b}}\\=2 \sqrt{a b} \cdot c

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question