Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 18 maths textbook solution

Explain Solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Exercise Fill in the blanks Question 18 maths textbook solution    

Answers (1)

Answer:

              75

Hint:

For maxima or minima we must have {f}'\left ( x \right )=0

Given:

\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}

Solution:

Let

\begin{aligned} &f^{\prime}(x)=\left(x^{2}-\frac{250}{x^{2}}\right) \\ \end{aligned}

Then

\begin{aligned} &f^{\prime}(x)=\left(2 x-\frac{250}{x^{2}}\right) \\ &f^{\prime \prime}(x)=\left(2+\frac{500}{x^{3}}\right) \\ &f^{\prime}(x)=0 \Rightarrow 2 x^{3}-250=0 \\ &2 x^{3}=250 \\ &x^{3}=125 \\ &x=5 \end{aligned}

\begin{aligned} &{{f}'}'(5)=\left(2+\frac{500}{125}\right)=6> 0 \\ \end{aligned} 

\therefore f\left ( x \right ) is minimum at x=5 and minimum value

=\left ( 25+\frac{250}{x} \right )=75

Posted by

Infoexpert

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 Board exams twice a year from 2025, no plan for semesters: MoE
anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question