A wire of length $36$ m is cut into two pieces, one of the pieces 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 $k$ (meter), then $(\frac{4}{π} + 1) k$ is equal to

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Solution

Let $x$ m is used to form circle and $(36 - x)$ m is used to form square.
If radius of circle is r
$2 π r = x \Rightarrow r = \frac{x}{2 π}$
Side of Square $= \frac{36 - x}{4}$
Sum of Area $= π {(\frac{x}{2 π})}^{2} + {(\frac{36 - x}{4})}^{2} = A$
$\frac{d A}{d x} = \frac{2 x}{4 π} - \frac{2}{4} (\frac{36 - x}{4})$

To minimize $\frac{d A}{d x} = 0 \Rightarrow x = \frac{36 π}{4 + π} = k$
$(\frac{4}{π} + 1) \cdot \frac{36 π}{4 + π} = \frac{4 + π}{π} \cdot \frac{36 π}{4 + π} = 36$

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A wire of length 36 m is cut into two pieces, one of the pieces 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 (meter), then (4π+1)k is equal to

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A wire of length $36$ m is cut into two pieces, one of the pieces 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 $k$ (meter), then $(\frac{4}{π} + 1) k$ is equal to