A wire of length 36 text{ m} is cut into two | vedclass

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(B) Let the length of the wire used for the square be $x$ and the length used for the circle be $y$. Then $x + y = 36$,so $y = 36 - x$.The side of the

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$36$

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(B) Let the length of the wire used for the square be $x$ and the length used for the circle be $y$. Then $x + y = 36$,so $y = 36 - x$.The side of the square is $s = \frac{x}{4}$,so the area of the square is $A_1 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}$.The circumference of the circle is $y = 2\pi r$,so the radius is $r = \frac{y}{2\pi} = \frac{36-x}{2\pi}$. The area of the circle is $A_2 = \pi r^2 = \pi \left(\frac{36-x}{2\pi}\right)^2 = \frac{(36-x)^2}{4\pi}$.The total area is $A(x) = \frac{x^2}{16} + \frac{(36-x)^2}{4\pi}$.To find the minimum,we differentiate with respect to $x$: $A'(x) = \frac{2x}{16} + \frac{2(36-x)(-1)}{4\pi} = \frac{x}{8} - \frac{36-x}{2\pi}$.Setting $A'(x) = 0$,we get $\frac{x}{8} = \frac{36-x}{2\pi} \Rightarrow \pi x = 4(36-x) \Rightarrow \pi x = 144 - 4x \Rightarrow x(\pi + 4) = 144 \Rightarrow x = \frac{144}{\pi + 4}$.The circumference of the circle is $k = y = 36 - x = 36 - \frac{144}{\pi + 4} = \frac{36\pi + 144 - 144}{\pi + 4} = \frac{36\pi}{\pi + 4}$.We need to find $\left(\frac{4}{\pi} + 1\right) k = \left(\frac{4+\pi}{\pi}\right) \left(\frac{36\pi}{\pi + 4}\right) = 36$.

$A$ wire of length $36 \text{ m}$ is cut into two pieces. One piece is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum,and the circumference of the circle is $k \text{ m}$,then $\left(\frac{4}{\pi}+1\right) k$ is equal to ..... .

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