A wire of length 2 units is cut into two parts which are bent respectively to form a square of side $=x$ units and a circle of radius $=r$ units. If the sum of the areas of the square and the circle formed is minimum, then: $$(1) (4-π)x = πr$$

$$(2) x = 2r$$

$$(3) 2x = r$$

$$(4) 2x = (π + 4)r$$

(1), (2), (3), (4) all are the options

I tried using hit and trial and got the answer as option B however I would like to solve it in a formulic way.

  • $\begingroup$ what have you tried? $\endgroup$ – 5201314 Jan 24 at 11:54
  • $\begingroup$ @5201314 I have now written in question what I tried $\endgroup$ – Ram Sale Jan 24 at 12:43
  • $\begingroup$ Do you know the equations for the perimeter and area of a square and circle? $\endgroup$ – K.defaoite Feb 7 at 17:02

The total of their perimeters is $4x + 2\pi r = 2 \implies r = \frac{1-2x}{\pi}$

Total area $A = x^2 + \pi r^2 = x^2 + \frac{(1-2x)^2}{\pi}$

$ = \frac{4+\pi}{\pi} [x^2 - \frac{4}{4+\pi}x + \frac{1}{4+\pi}]$

$ = \frac{4+\pi}{\pi} [(x - \frac{2}{4+\pi})^2 + \frac{\pi}{(4+\pi)^2}]$

That clearly shows that the total area is minimum when $x = \frac{2}{4+\pi}$, so $r = \frac{1}{4+\pi}$. Also the minimum total area is $\frac{1}{4+\pi}$.

  • $\begingroup$ Can you please specify which option you are referring to as your answer is quite confusing to me $\endgroup$ – Ram Sale Jan 24 at 13:30
  • $\begingroup$ You need to tell me what is confusing so I can explain. If you see the last line, I have written the value of $x$ and $r$. That shows $x = 2r$. $\endgroup$ – Math Lover Jan 24 at 13:33
  • $\begingroup$ Oh sorry being silly from my end. Thanks for the answer $\endgroup$ – Ram Sale Jan 24 at 13:42
  • $\begingroup$ You are welcome! $\endgroup$ – Math Lover Jan 24 at 13:52

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