Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 38.

Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 38.

Answers (1)

h =\frac{l}{2}

Hint: For maxima and minima value of S, we must have \frac{dS}{dl}=0

Given: Let l,h,v and S be the length, height, volume and surface area of the tank constructed

Since volume, v is constant  l^2h=v\Rightarrow h=\frac{v}{l^2 }                       →i

Solution:

Let S denote the total surface area, then we have S=l^2+4lh

\Rightarrow S=l^2+4l\times \frac{v}{l^2 }                             →fromi

\Rightarrow S=l^2+\frac{4v}{l}                              …(ii)

Differentiating S w.r.t 'l', we get

\frac{dS}{dl}= \frac{d}{dl}\left (l^2+\frac{4v}{l} \right )

\begin{aligned} &\Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+\frac{d}{d l}\left(\frac{4 v}{l}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(\frac{1}{2}\right) \quad\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ \Rightarrow \frac{d S}{d l}=\frac{d}{d l}\left(l^{2}\right)+4 v \frac{d}{d l}\left(l^{-1}\right) \end{aligned} \\ &\Rightarrow \frac{d S}{d l}=2 l^{2-1}+4 v\left(-1 l^{-1-1}\right) \quad\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \end{aligned}

\Rightarrow \frac{dS}{dl}=2l-\frac{4v}{l^2 }                                    …(iii)

For maximum or minimum of S, we have

\Rightarrow \frac{dS}{dl}=0

\Rightarrow 2l-\frac{4v}{l^2 }=0                                 [using (iii)]⇒\frac{2l^3-4v}{l^2}=0]

\Rightarrow 2l^3-4v=0

\Rightarrow 2l^3=4v\Rightarrow l^3=2v                             …(iv)

Again, differentiating S w.r.t 'l', then \frac{d^{2} S}{d l^{2}}=\frac{d}{d l}\left(\frac{d S}{d l}\right)=\frac{d}{d l}\left(2 l-\frac{4 v}{l^{2}}\right)                     [using (iii)]

\begin{aligned} &\Rightarrow \frac{d^{2} S}{d l^{2}}=\frac{d}{d l}(2 l)+\frac{d}{d l}\left(\frac{-4 v}{l^{2}}\right) \quad\left[\because \frac{d}{d x}\left\{(f(x)+g(x)\}=\frac{d}{d x}(f(x))+\frac{d}{d x}(g(x))\right]\right. \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}(l)-4 v \frac{d}{d l}\left(\frac{1}{l^{2}}\right) &\left[\because \frac{d}{d x}\left(a x^{n}\right)=a \frac{d}{d x}\left(x^{n}\right)\right] \\ & \Rightarrow \frac{d^{2} S}{d l^{2}}=2 \frac{d}{d l}\left(l^{1}\right)-4 v \frac{d}{d l}\left(l^{-2}\right) \end{aligned} \end{aligned}

\begin{aligned} &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}}=2\left(1 l^{1-1}\right)-4 v\left(-2 l^{-2-1}\right) &\left[\because \frac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right] \\ \Rightarrow \frac{d^{2} S}{d l^{2}}=2(1)+8 v\left(l^{-3}\right) \end{aligned} \\ &\begin{aligned} \Rightarrow \frac{d^{2} S}{d l^{2}} &=2+\frac{8 v}{l^{3}} \\ &=2+\frac{8}{2} \\ &=2+4 \end{aligned} \end{aligned}

                =6>0

     

\therefore \frac{d^2S}{dl^2}>0 Hence the surface area is minimum, h = \frac{v}{l^2} Substitute the value of v=\frac{l^3}{2}  from (iv) in eqni, we get \frac{l^{3}}{2 l^{2}} \Rightarrow h=\frac{l}{2}

Hence proved.

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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question