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. $2x=(\pi +4)r$
• B. $(4-\pi )x=\pi r$
• C. $x=2r$
• D. $2x=r$

Answer 1

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, $x=\frac{1-\pi r}{2}$

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

$\frac{dA}{dr}=\frac{2}{4}(1-\pi r)\times (-\pi )+2\pi r=0$ for minimum.

$\Rightarrow \pi r-1+4r=0$

$\Rightarrow r(4+\pi )=1$

$\Rightarrow r=\frac{1}{4+\pi }$

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

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

$\therefore x=2r$

(Since only one value of r obtain. Hence verification is not done for Minima it would be Minima there)