Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 26 maths.

Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 26 maths.

Answers (1)

r = 7cm

Hint: For maxima or minima value of s we must have \frac{ds}{dr}=0

Given: Let the height, radius of base and surface area of cylinder be h,r,v.

Solution:

Volume =\pi r^2 h

2156=\pi r^2 h

2156=\frac{22}{7} r^2 h

h = \frac{2156 \times 7}{22r^2},h = \frac{686}{r^2}                          .........(1)

Surface area = 2\pi rh+\pi r^2h 

=\frac{4132}{r}+\frac{44}{r^2}                 ......from (1)

\begin{aligned} &\frac{d s}{d r}=\frac{4312}{-r^{2}}+\frac{88 r}{7} \\ &\frac{d s}{d r}=0 \\ &-\frac{4312}{r}+\frac{88 r}{7}=0 \\ &\frac{4312}{r}=\frac{88 r}{7} \Rightarrow r^{3}=\frac{4321 \times 7}{88} \\ &r^{3}=343, r=7 \mathrm{~cm} \end{aligned}

Now,

\begin{aligned} &\frac{d^{2} s}{d r^{2}}=\frac{8624}{r^{3}}+\frac{88}{7} \\ &=\frac{8624}{343}+\frac{88}{7}=\frac{176}{7}>0 \end{aligned}

So, the surface area is maximum when r = 7

 

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 Single Girl Child Scholarship 2024: Registration extended till January 10; eligibility criteria, amount
anam-khan

CBSE wants international boards reined in; letter to education ministry seeks directions for AIU
anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

Latest Question