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A wire, 4 meters long, is cut into two pieces. One is bent into a shape pf a square and the other into a shape of a circle.

1. If the length of wire used to make the circle is x metres, write in terms of $x$ the length of the sides of the square in metres.
2. Show the sum of the areas of the circle and the square is given by $f(x) = (\frac{1}{16} + \frac{1}{4\pi})x^{2} - \frac{x}{2} + 1$
3. How should the wire be cut so that the sum of the areas of the circle and the square is a minimum?
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1. $= \dfrac{4-x}{4}$m

2. Area of square $=(\dfrac{4-x}{4})^2$

Area of circle $= \pi. r^2$

but circumference $= \pi.d$

$x= \pi.d$

$d= \dfrac{x}{\pi}$

$r= \dfrac{x}{2}.\pi$

$A = \dfrac{(\pi.x^2)}{4}\pi^2 = \dfrac{x^2}{4\pi}$

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

$= (\dfrac{1}{16} + \dfrac{1}{4\pi}).x^2 - \dfrac{x}{2} + 1$

3. $\dfrac{dA}{dx} = 2.x.(\dfrac{1}{16} + \dfrac{1}{4\pi}) - \dfrac{1}{2} = 0$

, solving $x= 3.52$m for circle and $4-3.52= 0.48$m for square
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