JEE MAIN - Mathematics (2016 (Offline) - No. 11)
A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then:
$$x=2r$$
$$2x=r$$
$$2x = \left( {\pi + 4} \right)r$$
$$\left( {4 - \pi } \right)x = \pi \,\, r$$
Explanation
$$4x + 2\pi r = 2$$ $$\,\,\,$$ $$ \Rightarrow 2x + \pi r = 1$$
$$S = {x^2} + \pi {r^2}$$
$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$
$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$
$$ \Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$
$$ \Rightarrow r = {1 \over {\pi + 4}}$$
$$ \Rightarrow x = {2 \over {\pi + 4}}\,$$
$$ \Rightarrow x = 2r$$
$$S = {x^2} + \pi {r^2}$$
$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$
$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$
$$ \Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$
$$ \Rightarrow r = {1 \over {\pi + 4}}$$
$$ \Rightarrow x = {2 \over {\pi + 4}}\,$$
$$ \Rightarrow x = 2r$$
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