You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece w

Question

You have a wire that is 20 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum

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Amelia 1 week 2021-09-15T19:13:13+00:00 1 Answer 0

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    2021-09-15T19:14:22+00:00

    Answer:

    Therefore the circumference of the circle is =\frac{20\pi}{4+\pi}

    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

    \Rightarrow s=\frac{10-\pi r}{2}

    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

    A=\pi r^2+ (\frac{10-\pi r}{2})^2

    \Rightarrow A= \pi r^2+(\frac{10}{2})^2-2.\frac{10}{2}.\frac{\pi r}{2}+ (\frac{\pi r}{2})^2

    \Rightarrow A=\pi r^2 +25-5 \pi r +\frac{\pi^2r^2}{4}

    \Rightarrow A=\pi r^2\frac{4+\pi}{4} -5\pi r +25

    For maximum or minimum \frac{dA}{dr}=0

    Differentiating with respect to r

    \frac{dA}{dr}= \frac{2\pi r(4+\pi)}{4} -5\pi

    Again differentiating with respect to r

    \frac{d^2A}{dr^2}=\frac{2\pi (4+\pi)}{4}    > 0

    For maximum or minimum

    \frac{dA}{dr}=0

    \Rightarrow \frac{2\pi r(4+\pi)}{4} -5\pi=0

    \Rightarrow r = \frac{10\pi }{\pi(4+\pi)}

    \Rightarrow r=\frac{10}{4+\pi}

    \frac{d^2A}{dr^2}|_{ r=\frac{10}{4+\pi}}=\frac{2\pi (4+\pi)}{4}>0

    Therefore at r=\frac{10}{4+\pi}  , A is minimum.

    Therefore the circumference of the circle is

    =2 \pi \frac{10}{4+\pi}

    =\frac{20\pi}{4+\pi}

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