Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 6 sub question (ii) maths textbook solution.

Please solve RD Sharma class 12 Chapter maxima and minima exercise 17.5 question 6 sub question (ii) maths textbook solution.

Answers (1)

\frac{L}{\sqrt{3}}

Hint: For maximum or minimum value of we must have \frac{dm}{dx}=0

Given: m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}

Solution: m=\frac{Wx}{3}-\frac{Wx^3}{3L^2}

\begin{aligned} &\frac{d m}{d x}=\frac{W}{3}-3 \times \frac{W x^{2}}{3 L^{2}} \\ &=\frac{W}{3}-\frac{W x^{2}}{L^{2}} \end{aligned}

For maximum or minimum,

\frac{dm}{dx}=0

\begin{aligned} &\frac{W}{3}-\frac{W x^{2}}{L^{2}}=0 \\ &\frac{W}{3}=\frac{W x^{2}}{L^{2}} \\ &x=\frac{L}{\sqrt{3}} \end{aligned}

\frac{d^{2} m}{d x^{2}}=\frac{-2 W x}{L^{2}}<0

So m is minimum at  x=\frac{L}{\sqrt{3}}

Posted by

infoexpert24

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 asks schools to ensure error-free LOC forms for Classes 10, 12; edit window from October 13
anam-khan

‘You ruined my career’: Students hit back at CBSE over 'abrupt' removal of additional subjects
anam-khan

‘Careers are at stake’: CBSE removes additional subject option for private students without prior notice

Latest Question