Question

A wire of length $28$cm is to be cut into $2$ pieces, $1$ piece is to be made into a square & the other one into a circle what should be the length of $2$ piece so that the combined area of the square & the circle is min?

Answer 1

Let, one part be of length x, then the other part $=28-x$

let, the part with length x is converted to circle.

$\begin{array}{c}2\pi r=x\\ r=\frac{x}{2}\pi \\ Areaofcircle=\pi r^2=\pi {\left(\frac{x}{4\pi }\right)}^2\\ =\frac{x^2}{4\pi }\end{array}$

Second is converted to square

$\begin{array}{c}sideofsquare=\frac{28-x}{4}\\ Area={\left(side\right)}^2={\left(\frac{28-x}{4}\right)}^2\\ TotalArea,A=\frac{x^2}{4\pi }+{\left(\frac{28-x}{4}\right)}^2\\ \frac{dA}{dx}=\frac{2x}{4\pi }+\frac{2}{16}\left(28-x\right)\left(-1\right)\\ =\frac{x}{2\pi }-\frac{28-x}{8}\\ put,\frac{dA}{dx}=0\\ \frac{x}{2\pi }=\frac{28-x}{8}\\ 4x=28\pi -\pi x\\ 4x+\pi x=28\pi \\ x=\frac{28\pi }{4+\pi }\\ otherpart:-\\ 28-x=28-\frac{28\pi }{4+\pi }=\frac{112}{4+\pi }\\ \frac{d^{2}A}{d{x}^2}=\left(\frac{1}{2\pi }+\frac{1}{8}\right):-+ve\\ \therefore Aisminimum\\ Whenx=\frac{28\pi }{4+\pi }and\left(28-x\right)=\frac{112}{4+x}\end{array}$