A wire of length 28 , m is to be cut into two | vedclass

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(A) Let the length of the wire used for the square be $l \, m$. Then the length of the wire used for the circle is $(28-l) \, m$.The side of the squar

Vedclass · Accepted Answer

Square: $\frac{112}{\pi+4} \, m$,Circle: $\frac{28\pi}{\pi+4} \, m$

Vedclass · Answer

(A) Let the length of the wire used for the square be $l \, m$. Then the length of the wire used for the circle is $(28-l) \, m$.The side of the square is $s = \frac{l}{4}$. The area of the square is $A_s = s^2 = \frac{l^2}{16}$.For the circle,the circumference is $2\pi r = 28-l$,so the radius is $r = \frac{28-l}{2\pi}$. The area of the circle is $A_c = \pi r^2 = \pi \left( \frac{28-l}{2\pi} \right)^2 = \frac{(28-l)^2}{4\pi}$.The total area $A = A_s + A_c = \frac{l^2}{16} + \frac{(28-l)^2}{4\pi}$.To find the minimum,we differentiate $A$ with respect to $l$: $\frac{dA}{dl} = \frac{2l}{16} + \frac{2(28-l)(-1)}{4\pi} = \frac{l}{8} - \frac{28-l}{2\pi}$.Setting $\frac{dA}{dl} = 0$,we get $\frac{l}{8} = \frac{28-l}{2\pi} \Rightarrow \pi l = 4(28-l) \Rightarrow \pi l = 112 - 4l \Rightarrow l(\pi+4) = 112 \Rightarrow l = \frac{112}{\pi+4}$.The second derivative $\frac{d^2A}{dl^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$,confirming a minimum.The length for the square is $\frac{112}{\pi+4} \, m$ and the length for the circle is $28 - \frac{112}{\pi+4} = \frac{28\pi+112-112}{\pi+4} = \frac{28\pi}{\pi+4} \, m$.

$A$ 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?

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