Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Show that of all the rectangles with a given area the one with smallest perimeter is a square.

Can you answer this question?

1 Answer

0 votes
  • Second derivative test: Suppose $f$ is continuous on an open interval that contains
  • (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
  • (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $c$
Step 1:
Let the given area be A and x, $\large\frac{A}{x}$ the sides of a, rectangle (Since area= length x breadth, the length and breadth are $x, \large\frac{A}{x}$)
Step 2:
The perimeter of the rectangle =2 (length+breadth)
$p=2 (x +\frac{A}{x})$
Step 3:
At extreme values of $p , \large\frac{dp}{dx}$$=0$
$\therefore 2(1-\large\frac{A}{x^2})$$=0$$=> x= A^{1/2}$
Now $\large\frac{d^2p}{dx^2}$$= 2 \times \large\frac{2A}{x^3}=\large\frac{4A}{x^3}$
$\large\frac{d^2p}{dx^2(x=A^{1/2})}$$= \large\frac{4A}{A^{3/2}}=\large\frac{4}{\sqrt A}$$ > 0$
$\therefore x=A^{1/2}$ corresponds to a minimum value of P
Step 4:
The sides of the rectangle of minimum perimeter are $ A^{1/2},A^{1/2}$ =>it is the square
answered Aug 1, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App