Optimization problem

Nov 2017
A piece of wire 202020 m long is cut into two pieces, the length of the first piece being x m. is cut into two pieces. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.

could someone please help me out with this one? any insight is appreciated


MHF Helper
Nov 2013
piece 1 is $x$ long and is bent into a circle with circumference $x$

The radius of this circle is $r = \dfrac{x}{2\pi}$ and the area is $\pi r^2 = \pi \dfrac{x^2}{4\pi^2} = \dfrac{x^2}{4\pi}$

let $L=202020$

The second piece of wire is $L-x$ long and is bent into a rectangle of dimensions $2w \times w$

The perimeter of this rectangle is $L-x = 2(2w+w)=6w$

$w = \dfrac{L-x}{6}$

The area of this rectangle is $(2w)\times w = 2w^2 = \dfrac {1}{18} (L-x)^2$

So the total area is given by

$A = \dfrac{x^2}{4\pi}+\dfrac{1}{18} (L-x)^2$