Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

10) The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answers (1)

best_answer

It is given that
the sum of the perimeter of a circle and square is k = 2\pi r + 4a = k\Rightarrow a = \frac{k - 2\pi r}{4} 
Let the sum of the area of a circle and square(A) = \pi r^2 + a^2
A = \pi r^2 + (\frac{k-2\pi r}{4})^2
A^{'}(r) = 2\pi r + 2(\frac{k-2\pi r}{16})(- 2\pi)\\ A^{'}(r) = 0\\ 2\pi (\frac{8r-k-2\pi r}{8}) = 0\\ r = \frac{k}{8-2\pi}
Now,
A^{''}(r) = 2\pi (\frac{8-2\pi }{8}) = 0\\ A^{''}(\frac{k}{8-2\pi}) > 0
Hence, r= \frac{k}{8-2\pi} is the point of minima
a = \frac{k-2\pi r}{4} = \frac{k-2\pi \frac{k}{8-2\pi}}{4}=2 \frac{k}{8-2\pi} = 2r
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle

Posted by

Gautam harsolia

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 2025 from February 15; Class 10, 12 subject-wise marks distribution
anam-khan

CBSE Date Sheet 2025: Class 10, 12 board exam dates soon on cbse.gov.in; previous trends, sample papers
anam-khan

CBSE Board Exams 2025: 75% attendance must for students to be eligible, reiterates board

Latest Question