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 $\left( {{4 \over \pi } + 1} \right)k$ is equal to _____________.
Answer (integer)
36
Solution
Let x + y = 36<br><br>x is perimeter of square and y is perimeter of circle side of square = x/4<br><br>radius of circle = ${y \over {2\pi }}$<br><br>Sum Areas = ${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$<br><br>$= {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$<br><br>For min Area :<br><br>$x = {{144} \over {\pi + 4}}$<br><br>$\Rightarrow$ Radius = y = 36 $-$ ${{144} \over {\pi + 4}}$<br><br>$\Rightarrow$ k = ${{36\pi } \over {\pi + 4}}$<br><br>$\left( {{4 \over \pi } + 1} \right)k$ = 36
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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