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

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  • 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 $x,y $ be the length and breadth of a rectangle. Whose permeter is the given value P
$\therefore\; 2(x+y)=p=>y=\large\frac{p}{2}$$-x$
The sides of the rectangle are $ x, \large\frac{p}{2}$$-x$
Step 2:
The area of the rectangle is
$A=x \bigg (\large \frac{p}{2}$$-x\bigg)=\large\frac{px}{2}$$-x^2$
Step 3:
At extreme values of $A, \large\frac{dA}{dx}$$=0$
$\large\frac{d^2p}{dx^2}$$=-2 <0=>x =\large\frac{p}{4}$
Corresponds to a maximum value of A.
$\therefore$ the area is maximum when the sides are $\large\frac{p}{4},\frac{p}{4}$ .
The sides are equal => this is a square
answered Aug 5, 2013 by meena.p

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