Question

hey guys here is your question ➡

➡A wire of length 2 unit is cut into two parts which are bent respectively to form a square of side = x unit and a circle of radius = r unit , if the sum of the area of the square and the circle so formed is minimum , then:-​

Answer 1

Answer:

The answer is 2r

Step-by-step explanation:

Let length of two parts be ‘a’ and ‘2 - a’

As per condition given, we write

a=4xand2−a=2πr

x=a4 and r=2−a2π

∴A(square)=(a4)2=a216 and

A(circle)=π[(2−a)2π]2=π(4+a2−4a)4π2

=(a2−4a+4)4π

f(a)=a216+a2−4a+44π

∴f(a)=a2π+4a2−16a+1616π

f′(a)=116π[2aπ+8a−16]

f′(a)=0=>2aπ+8a−16=0

=> 2aπ+8a=16

x=a4=2π+4

and r=2−a2π

=2−8π+42π

=2π+8−82π(π+4)

=1π+4

x=2π+4 and r=1π+4

x=2r

Answer 2

Answer:

x = 2r

Step-by-step explanation:

Given, Side of square = x units and radius of circle = r units.

Given, Length of wire = 2 units.

Perimeter of Square + Perimeter of circle = 2

∴ 4x + 2πr = 2

⇒ 2x + πr = 1

⇒ r = (1 - 2x)/π

(i)

S = x² + πr²

  = x² + π(1 - 2x/π)²

  = x² + (1 - 2x)/π

Therefore,

dS/dx = 2x + (1/π)2(1 - 2x)(-2)

          = 2x - (4/π)(1 - 2x) = 0

Hence,

⇒ d²S/dx² = 2x + (4/π) (2) > 0

Now,

⇒ x - (2/π)(1 - 2x) = 0

⇒ πx - 2 + 4x = 0

⇒ x = (2/π + 4)

⇒ x = 2r {∵ πr => 1 - 4/π+r = π/π+4 => r = 1/π + 4}

Therefore, x = 2r.

Hope it helps!