Respuesta :
Answer:
Therefore the circumference of the circle is [tex]=\frac{20\pi}{4+\pi}[/tex]
Step-by-step explanation:
Let the side of the square be s
and the radius of the circle be r
The perimeter of the square is = 4s
The circumference of the circle is =2πr
Given that the length of the wire is 20 cm.
According to the problem,
4s + 2πr =20
⇒2s+πr =10
[tex]\Rightarrow s=\frac{10-\pi r}{2}[/tex]
The area of the circle is = πr²
The area of the square is = s²
A represent the total area of the square and circle.
A=πr²+s²
Putting the value of s
[tex]A=\pi r^2+ (\frac{10-\pi r}{2})^2[/tex]
[tex]\Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2[/tex]
[tex]\Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}[/tex]
[tex]\Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25[/tex]
For maximum or minimum [tex]\frac{dA}{dr}=0[/tex]
Differentiating with respect to r
[tex]\frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi[/tex]
Again differentiating with respect to r
[tex]\frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}[/tex] > 0
For maximum or minimum
[tex]\frac{dA}{dr}=0[/tex]
[tex]\Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0[/tex]
[tex]\Rightarrow r = \frac{10\pi }{\pi(4+\pi)}[/tex]
[tex]\Rightarrow r=\frac{10}{4+\pi}[/tex]
[tex]\frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0[/tex]
Therefore at [tex]r=\frac{10}{4+\pi}[/tex] , A is minimum.
Therefore the circumference of the circle is
[tex]=2 \pi \frac{10}{4+\pi}[/tex]
[tex]=\frac{20\pi}{4+\pi}[/tex]
