Application of differentation

Let length of side of square = x. Therefore, length of wire for bending to a square = 4x. And length for bending to a circle =( 6-4x)= circumference of circle. Therefore, radius of circle = (6-4x)/2p, where p = pi.
Area of square = x^2.
Area of circle = p[(6-4x)/2p]^2 = (6-4x)^2/4p = (3-2x)^2/p.
Therefore, total area, A= x^2 + (3-2x)^2/p
dA/dx = 2x -4(3-2x)/p.
Put dA/dx = 0, we get 2x = 4(3-2x)/p
px = 6-4x
x= 6/(4+p).
Therefore, area is a maximum when x = 6/(4+pi).

2008-07-24 15:20:54 補充：
Correction: Sorry, x=6/(4+pi) should be a minimum because the coeff. of x^2 is postive. So max. happens at the end points, that is x=0 or x=6/4. When x = 0, area = 9/pi, when x = 6/4, area = 2.25. Therefore, max. area happens when x =0.

Let x be the length of the square

4x + 2丌r = 6
r = (3-2x)/丌

Area of the square is
x^2

Area of the circle is
丌[(3-2x)/丌]^2
= (3-2x)^2 / 丌

Sum of the areas of the them is
S = (3-2x)^2 / 丌 + x^2
dS/dx = 2(-2)(3-2x)/丌 + 2x

When dS/dx = 0
-4(3-2x)/丌 + 2x = 0
2丌x = 4(3-2x)
丌x = 6 - 4x
x = 6/(丌+4)
x = 0.8401 cm //

* d^2S/dx^2 = 8/丌 + 2 > 0
S = (3-2( 6/(丌+4))^2 / 丌 + [6/(丌+4) ]^2 = 0.7621

when x = 0.8401 , S is minimum
Now , when x = 0 ,
S = (3)^2 / 丌 = 2.8648 cm^2

when x = 6/4
S = 2.25

S is maximum when x = 0