# How to solve the problems in which wire is cut into 2 parts [closed]

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.

• what have you tried? – 5201314 Jan 24 at 11:54
• @5201314 I have now written in question what I tried – Ram Sale Jan 24 at 12:43
• Do you know the equations for the perimeter and area of a square and circle? – 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 $$\frac{1}{4+\pi}$$.

• Can you please specify which option you are referring to as your answer is quite confusing to me – Ram Sale Jan 24 at 13:30
• 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 $x = 2r$. – Math Lover Jan 24 at 13:33
• Oh sorry being silly from my end. Thanks for the answer – Ram Sale Jan 24 at 13:42
• You are welcome! – Math Lover Jan 24 at 13:52