Question

A wire of length 28 cm is to be into two pieces. One of the pieces to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Answer 1

Let the length of the piece bent into the shape of a circle be x (units) so that the length of the other piece bent into the shape of a square is (28-x) unit.s

Now radius of the circle $=\frac{Circumference}{2\pi }=\frac{x}{2\pi }$
Area of the circle $=\pi (radius{)}^2=\pi {\left(\frac{x}{2\pi }\right)}^2=\frac{x^2}{4\pi }$
Also, the length of each side of the square $=\frac{Perimeter}{4}=\frac{28-x}{4}$
Area of the square $=(Side{)}^2={\left(\frac{28-x}{4}\right)}^2=\frac{(28-x{)}^2}{16}$
Let f(x) be the sum of the areas of the two figures, then
$F\left(x\right)=\frac{x^2}{4\pi }+\frac{(28-x{)}^2}{16}\dots \left(i\right)$
On differentiating Eq. (i) twice w.r.t. x, we get
$f^{\prime }\left(x\right)=\frac{2x}{4\pi }+\frac{2(28-x)(-1)}{16}=\frac{x}{2\pi }-\frac{28-x}{8}andf"\left(x\right)=\frac{1}{2\pi }-\frac{(-1)}{8}=\frac{1}{2\pi }+\frac{1}{8}$
Now put $f^{\prime }\left(x\right)=0\Rightarrow \frac{x}{2\pi }-\frac{28-x}{8}=0\Rightarrow 4x-\pi (28-x)=0$
$\Rightarrow 4x+\pi x-28\pi =0\Rightarrow x(4+\pi )=28\pi \Rightarrow x=\frac{28\pi }{4+\pi }$
and for this value of $f"\left(x\right)=\frac{1}{2\pi }+\frac{1}{8}>0$
$\therefore$ f(x) has a local minima at $x=\frac{28\pi }{4+\pi }$
Since, f is continuous in (0,28) and has only one extreme point at $\frac{28\pi }{4+\pi }ϵ(0,28)$, therefore, f(x) is absolutely minimum for $x=\frac{28\pi }{4+\pi }$
Hence, the length of circular piece = $\frac{28\pi }{4+\pi }m.$
and square piece $=28-\frac{28\pi }{4+\pi }=\frac{112}{4+\pi }$.