A wire of length 2 units is cut into two part | vedclass

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(A) Given that the total length of the wire is $2$ units.Perimeter of the square $= 4x$ and circumference of the circle $= 2\pi r$.Thus,$4x + 2\pi r =

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$x = 2r$

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(A) Given that the total length of the wire is $2$ units.Perimeter of the square $= 4x$ and circumference of the circle $= 2\pi r$.Thus,$4x + 2\pi r = 2$,which simplifies to $2x + \pi r = 1$.From this,we get $r = \frac{1 - 2x}{\pi}$.The sum of the areas $A$ is given by $A = x^2 + \pi r^2$.Substituting the value of $r$,we get $A = x^2 + \pi \left( \frac{1 - 2x}{\pi} \right)^2 = x^2 + \frac{(1 - 2x)^2}{\pi}$.To find the minimum area,we differentiate $A$ with respect to $x$ and set it to zero:$\frac{dA}{dx} = 2x + \frac{2(1 - 2x)(-2)}{\pi} = 0$.$2x - \frac{4(1 - 2x)}{\pi} = 0$.Multiplying by $\pi$,we get $2\pi x - 4 + 8x = 0$.$x(2\pi + 8) = 4$,so $x = \frac{4}{2\pi + 8} = \frac{2}{\pi + 4}$.Now,substitute $x$ back into the perimeter equation: $2(\frac{2}{\pi + 4}) + \pi r = 1$.$\pi r = 1 - \frac{4}{\pi + 4} = \frac{\pi + 4 - 4}{\pi + 4} = \frac{\pi}{\pi + 4}$.Therefore,$r = \frac{1}{\pi + 4}$.Comparing $x$ and $r$,we see that $x = \frac{2}{\pi + 4}$ and $r = \frac{1}{\pi + 4}$,which implies $x = 2r$.

$A$ 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 so formed is minimum,then:

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