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

The correct option is C $$x = 2r$$Length of wire $$=2$$ unitsLength of side of square $$=x$$ units Radius of circle $$=r$$ unitsTotal 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

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 