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**helloying** A piece of wire of length 8cm is cut into two pieces,one of length x cm,the other of length (8-x)cm.The piece of lenght x cm is bent to form a circle with circumference x cm. The other piece is bent to form a square with perimeter (8-x)cm. Show that , as x varies, the sum of the areas enclosed by these two pieces of wire is a minimum when the radius of the circle is$\displaystyle \frac{4}{4+\pi}$

i found the area of the square to be $\displaystyle (4-\frac{x}{2})cm^2$

and the the radius of the circle is $\displaystyle \frac{x}{2\pi}$so the area of circle is $\displaystyle \frac{x^2}{4\pi}$

i add them up and i differentiate the eqn. i got $\displaystyle \frac{x}{2\pi}-0.5$

At min point,$\displaystyle \frac{x}{2\pi}-0.5$=0

thus x=$\displaystyle \frac{1}{\pi}$

then when i sub it into the radius, i did not get the ans. what is wrong?