Question

A piece of wire 40 cm long is to be cut into two pieces. One piece will be bent to form a circle and; the other will be bent to form a square. Find the lengths of the two pieces that cause the sum of the area of the circle and the area of the square to be a minimum. The perimeter of the square is cm The circumference of the circle is cm

Answer 1

The perimeter of circle  = 17.6 cm

Perimeter of square  = 22.4 cm

Step-by-step explanation:

The total length of the wire  = 40 cm

Let us assume the circumference of the circle  = k cm

So, the circumference of the square  = ( 40 - k) cm

Now,as circumference of circle = 2πR

⇒2πR = k cm    or, R  = $(\frac{k}{2\pi})$

Area  of the  circle = πR²  

or, A  = $\pi (\frac{k}{2\pi} )^2$  

 $= \pi \times (\frac{k^2}{(2\pi)^2} ) = \frac{k^2}{4 \pi} \\\implies A = \frac{k^2}{4 \pi}$   ....... (1)

Similarly , perimeter of square  = 4 x Side

⇒ 4 x Side  =  ( 40 - k) cm    ⇒ S = $(\frac{40 - k}{4} )$

Area  of the  square = (Side)²    = $(\frac{40 - k}{4} ) ^2$

 Solving, we get:  $A = \frac{1600 + k^2 - 80 k}{16}$   ....... (1)

So, combining (1) and (2), total area A is

$A = (\frac{k^2}{4 \pi} ) + \frac{k^2 + 1600 - 80 k }{16}$

or, we get: A  = 0.07962 k²  + 0.0625 k² + 100 - 5 k

A  =  0.1422 k² - 5 k + 100

For axis of symmetry :  $k = (\frac{-b}{2a} ) = (-\frac{-5}{2(0.1420}) = 17.6$

⇒ k = 17.6 inches

So, the perimeter of circle  = 17.6 cm

Perimeter of square  = (40 - 17.6 )  = 22.4 cm