Optimization problem

Nov 2017
1
0
alaska
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
 

romsek

MHF Helper
Nov 2013
6,748
3,037
California
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

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