Browse Questions

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

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}$