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

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Toolbox:
  • 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}$$=0$
$\large\frac{dp}{dx}$$=2(1-\large\frac{A}{x^2})$
$\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
 

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