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Question

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:


A
2x=(π+4) r
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B
(4π)x=π r
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C
x=2r
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D
2x=r
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Solution

The correct option is C $$x = 2r$$
Length of wire $$=2$$ units
Length of side of square $$=x$$ units 
Radius of circle $$=r$$ units
Total area $$=x^2+\pi r^2$$

Also, $$4x+2 \pi r=2$$
So, $$x=\dfrac {1- \pi r}{2}$$

Total area $$=\dfrac {(1-\pi r)^2}{4} + \pi r^2$$

$$\dfrac {dA}{dr} = \dfrac 24 (1-\pi r) \times (-\pi) + 2\pi r = 0$$ for minimum.

$$\Rightarrow \pi r-1+4r=0$$
$$\Rightarrow r(4+\pi)=1$$
$$\Rightarrow r=\dfrac {1}{4+\pi}$$

$$x=\dfrac {1-\pi r}{2}=\dfrac {1-\pi \left(\frac {1}{4+\pi}\right)}{2}$$

$$=\dfrac {4+\pi-\pi}{2(4+\pi)}=\dfrac {2}{4+\pi}$$

$$\therefore x=2r$$
(Since only one value of r obtain. Hence verification is not done for Minima it would be Minima there)

Applied Mathematics

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