Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Please Solve RD Sharma Class 12 Chapter 17 Maxima and Minima Exercise Case Study Based Questions question 5 subquestion (iv) Maths Textbook Solution.

Answers (1)

best_answer

Answer: $x = 2$

Hint: use the concept of maxima and minima.

Solution:

As we know,

$C^{\prime}(x)=3600\left(1-\frac{4}{x^{2}}\right)$

Putting $c^{\prime}(x)=0$

$\begin{aligned} 3600\left(1-\frac{4}{x^{2}}\right) &=0 . \\ \end{aligned}$

$1-\frac{4}{x^{2}} =0 \\$

$\frac{x^{2}-4}{x^{2}} =0 \\$

$x^{2}-4=0 \\$

$(x-2)(x+2) =0$

$\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

$\mathrm{c}^{\prime \prime}(\mathrm{x})=3600\left(0-4(-2) \mathrm{x}^{-2-1}\right) \\$

$\quad=3600(0-4(-2) \mathrm{x}^{-3} \\$

$=3600\left(8 \mathrm{x}^{-3}\right) \\$

$=28800 \mathrm{x}^{-3} \\$

$=\frac{28800}{\mathrm{x}^{3}}$

Putting $\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

$\begin{aligned} &c^{\prime \prime}(2)=\frac{28800>0}{(2)^{3}} \\ &c^{\prime \prime}(2)>0 \end{aligned}$

So minimum is $\begin{aligned} &\mathrm{x}=2 \\ \end{aligned}$

Posted by

infoexpert27

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 extends single girl child scholarship registration deadline till November 20

anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16

anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8

Latest Question