A wire of length $28{\rm{ m}}$ is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces so that the combined area of the square and the circle is minimum?

Let us assume that the length of the piece bent into the shape of a circle be $x{\rm{m}}$ and so length of the other piece bent into the shape of a square is $(28 - x){\rm{m}}$.

Circumference of circle is given by $= 2\pi r \Rightarrow 2\pi r = x \Rightarrow r = \cfrac{x}{{2\pi }}$
$\Rightarrow$ Area of the circle$= \pi {(r)^2} = \pi {\left( {\cfrac{x}{{2\pi }}} \right)^2} = \cfrac{{{x^2}}}{{4\pi }}$ .

Perimeter of square $= 4$(side)$\Rightarrow 28 - x = 4{\rm{(side)}}$
$\Rightarrow$ Side $= \cfrac{{28 - x}}{4}$

$\Rightarrow$ Area of the square $= {\left( {{\rm{side}}} \right)^2} = {\left( {\cfrac{{28 - x}}{4}} \right)^2} = \cfrac{{{{(28 - x)}^2}}}{{16}}$.

Let $A$ be the sum of the areas of the two shapes, then
$A = \cfrac{{{x^2}}}{{4\pi }} + \cfrac{{{{(28 - x)}^2}}}{{16}}$ … (i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{dA}}{{dx}} = \cfrac{{2x}}{{4\pi }} + \cfrac{{2(28 - x)( - 1)}}{{16}} = \cfrac{x}{{2\pi }} - \cfrac{{28 - x}}{8}$

To find the maximum or minimum area, $\cfrac{{dA}}{{dx}} = 0$
$\Rightarrow \cfrac{x}{{2\pi }} - \cfrac{{28 - x}}{8} = 0 \Rightarrow \cfrac{{4x - 28\pi + x\pi }}{{8\pi }} = 0$

$\Rightarrow 4x + x\pi = 28\pi = x = \cfrac{{28\pi }}{{4 + \pi }}$

$\cfrac{{{d^2}A}}{{d{x^2}}} = \cfrac{1}{{2\pi }} - \cfrac{{( - 1)}}{8} = \cfrac{1}{{2\pi }} + \cfrac{1}{8}$ and ${\left( {\cfrac{{{d^2}A}}{{d{x^2}}}} \right)_{x = \cfrac{{28\pi }}{{4 + \pi }}}} = \cfrac{1}{{2\pi }} + \cfrac{1}{8} > 0$

Hence, area $A$ is minimum at $x = \cfrac{{28\pi }}{{4 + \pi }}$

Therefore the wire must be cut at a distance of $\cfrac{{28\pi }}{{4 + \pi }}{\rm{m}}$ from one end.

Hence, the lengths of the two pieces are $28 - \cfrac{{28\pi }}{{4 + \pi }} = \cfrac{{112}}{{4 + \pi }}{\rm{m}}$ and $\cfrac{{28\pi }}{{4 + \pi }}{\rm{m}}$.