We have a 1 unit line and we can can cut it into 2 parts, where 1 part is used at the circumference of the square and the other as the circumference of a circle. How do i find out where to cut the line to get the least area as possible?

  • $\begingroup$ What have you tried? Can you write an equation expressing the total area, if the line was cut at $x$? $\endgroup$
    – Calvin Lin
    Apr 3 '20 at 10:43
  • $\begingroup$ cut up the wire and made x and 100-x as the cuts, then found A=A_s+A_c, derivated it and found 0,56 unit for the square, but how can i be sure that is correct? $\endgroup$ Apr 3 '20 at 10:50
  • $\begingroup$ Write it up (either in your question, or as an answer), and people can look it over. $\endgroup$
    – Calvin Lin
    Apr 3 '20 at 10:51
  • $\begingroup$ FWIW $x \approx 0.56$ is indeed the minimum. $\endgroup$
    – Calvin Lin
    Apr 3 '20 at 10:55
  • $\begingroup$ What? why edit it wrong? changed it now $\endgroup$ Apr 3 '20 at 11:08

Say the part of length $x$ forms the circle. That circle has radius $\frac x{2\pi}$ and area $\frac{x^2}{4\pi}$. The remaining part of length $1-x$ forms a square with side $\frac{1-x}4$ and area $\frac{(1-x)^2}{16}$. So we have to minimise $$\frac{(x-1)^2}{16}+\frac{x^2}{4\pi}=\left(\frac1{16}+\frac1{4\pi}\right)x^2-\frac x8+\frac1{16}$$ and the minimum occurs where the derivative is zero: $$\left(\frac18+\frac1{2\pi}\right)x-\frac18=0$$ $$x=\frac{1/8}{1/8+1/2\pi}=0.439900\dots$$ So this length is reserved for the circle, and the remaining part for the square.

  • $\begingroup$ This is what i did. Can this be used? $A=A_s+A_c$ such that $A(x)=(\frac{x}{4})^2+\pi(\frac{100-x}{2\pi})^2$, Then find the derivate: $A'(x)=2(\frac{x}{4})\cdot\frac{1}{4}+2\pi(\frac{100-x^2}{2\pi})(\frac{-1}{2\pi})\Longrightarrow$ $0=\frac{8}{x}+\frac{x-100}{2\pi}$, $0=\pi x+4(x-100)$, $0=4x+ \pi x-400\Longrightarrow$ $x=\frac{400}{4+\pi}\approx56$ $\endgroup$ Apr 3 '20 at 11:23
  • $\begingroup$ Yeah, i see that by looking at your solution! $\endgroup$ Apr 3 '20 at 11:25

Let the square and circle parts be partitioned $ (x, 1-x)$ respectively.

$$ x= 4a,\, 2 \pi r = (1-x),\, U= 4 a + 2 \pi r $$

$$ A = a^2 + \pi r^2 $$


simplifies to

$$ \frac{4 A}{\pi} = \frac{\pi x^2}{4}+(1-x)^2 = x^2(\pi/4+1)-2x+ 1 $$

By usual differentiation its parabola graph has at

$$ x=\frac{1}{1+\pi/4} \approx 0.44$$

gives a minimum for square/circle.

Also, by Lagrange Multiplier method we gain extra geometrical insight ( I mention this even if outside pre-calculus scope ):

$$ U= 4a+2\pi r;\, A = a^2+\pi r^2;$$

If each of $U$ and $A$ are partially differentiated with respect to $r$ and $a$ separately, and their ratios equated, we get a simple relation

$$ a = 2r$$

or the side should equal the circle diameter if laid side by side after dividing the given length of string as sketched below:

Square and Circle

  • $\begingroup$ $a^2=\frac{x^2}{16}$ not $\frac{x^2}{4}$ $\endgroup$ Apr 3 '20 at 12:14
  • $\begingroup$ answer should be 0.44 and 0.56 $\endgroup$ Apr 3 '20 at 12:15
  • 1
    $\begingroup$ Thanks @DanielMathias and@ AndersJensen Error is corrected. $\endgroup$
    – Narasimham
    Apr 3 '20 at 14:35

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