A wire of length 36m 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 $(\frac{π + 4}{π}) k$ is equal to

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answer is 36.

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Detailed Solution

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$2 πr = 36 - x \Rightarrow r = \frac{36 - x}{2 π}$

$S =$ sum of areas $= \frac{x^{2}}{16} + {πr}^{2}$

$S = \frac{x^{2}}{16} + π \frac{{(36 - x)}^{2}}{4 π^{2}} = \frac{x^{2}}{16} + \frac{{(36 - x)}^{2}}{4 π} S^{1} = \frac{2 x}{16} + \frac{2 (36 - x)}{4 π} (- 1)$

S is min or max $\Rightarrow S^{1} = 0$

$\frac{x}{16} = \frac{36 - x}{4 π} \Rightarrow πx = 144 - 4 x$

$\begin{array}{rcl} (π + 4) x & = & 144 \Rightarrow x = \frac{144}{π + 4} \\ 2 πr & = & k and r = \frac{36 - x}{2 π} \Rightarrow 2 πr = 36 - x \\ k & = & 36 - \frac{144}{π + 4} = \frac{36 π}{π + 4} \\ (\frac{π + 4}{π}) k & = & (\frac{π + 4}{π}) (\frac{36 π}{π + 4}) = 36 \end{array}$