Question

How should a wire $20$ cm long be divided into two parts, if one part is to be bent into a circle, the other part is to be bent into a square and the two plane figures are to have areas the sum of which is minimum? ($r$= radius and $x$ = side of square)

• A. $r=2x$
• B. $r=x$
• C. $2r=x$
• D. $3r=x$

Answer 1

The correct option is C $2r=x$

Let $x$ be the side of the square and $r$ the radius of the circle .

Given $P_1+P_2=20$

$4x+2\pi r=20$

$\Rightarrow 2x+\pi r=10$ ...(1)

Now, $A=A_1+A_2$

$\Rightarrow A=x^2+\pi r^2$

$={\left\{\frac{20-2\pi r}{4}\right\}}^2+\pi r^2$

or $A=\frac{1}{4}{(10-\pi r)}^2+\pi r^2$

Now,$\frac{dA}{dr}=\frac{1}{4}.2\{10-\pi r\}(-\pi )+2\pi r$

For maximum or minimum,

$\frac{dA}{dr}=0$

$\Rightarrow 4r=10-\pi r=2x$ by (1)

$\therefore 4r=2x$ or $x=2r$

i.e. the diameter of the circle is equal to side of the square.

Also $\frac{d^{2}A}{d{r}^2}=\frac{{\pi }^2}{2}+2\pi =+ive$

Hence, A is minimum.