A wire of length 28 m is to be cut into two pieces

Question: A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution: Let \mathrm{x} meters be the side of square and y meters be the radius of the circle.

A wire of length 28 m is to be cut into two pieces

Length of the wire = Perimeter of square + Circumference of circle

4 x+2 \pi y=28

\Rightarrow 2 x+\pi y=14

\Rightarrow y=\frac{14-2 x}{\pi}

\text { Area of square }=x^2, \text { and Area of circle }=\pi y^2

\text { Combined area }(\mathrm{A})=x^2+\pi y^2=x^2+\pi\left(\frac{14-2 x}{\pi}\right)^2

=x^2+\frac{4}{\pi}(7-x)^2

d\frac{dA}{d x}=2 x-\dfrac{8}{\pi}(7-x) \text { and }

\dfrac{d^2 A}{d x^2}=2+\dfrac{8}{\pi}

 Now For max and minima \dfrac{d A}{d x}=0

2 x-\dfrac{8}{\pi}(7-x)=0

\Rightarrow 2 x=\dfrac{8}{\pi}(7-x)

\Rightarrow 2 \pi x=56-8 x

\Rightarrow (2 \pi+8) x=56

\Rightarrow x=\dfrac{56}{2 \pi+8}=\dfrac{28}{\pi+4}

\text { And } \dfrac{d^2 A}{d x^2}=2+\dfrac{8}{\pi} \text { [Positive] }

\text { A is minimum when } x=\dfrac{28}{\pi+4}

Therefore, the length of one piece = 4 x=\dfrac{112}{\pi+4}

Length of second piece = 2\pi y= 28- \dfrac{28}{\pi+4}

= \dfrac{28\pi+112-112}{28+\pi}

= \dfrac{28\pi}{28+\pi}

 

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