# Wire of length 20m is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas? Round your answer to the nearest hundredth?

Jan 15, 2018

Application Of Derivatives

#### Explanation:

let side of square be x

therefore length of wire used for making the square is 4x

so the remaining 20-4x is used for making the circle

so circumference of circle = 20-4x

2$\pi$r = 20-4x

hence r = $\frac{10 - 2 x}{\pi}$

now let f(x) be sum of areas

therefore $f \left(x\right)$ = ${x}^{2} + \pi {r}^{2}$

= ${x}^{2} + {\left(10 - 2 x\right)}^{2} / \pi$ = ${x}^{2} + \frac{4 {x}^{2} - 40 x + 100}{\pi}$

now find derivative of $f \left(x\right)$ find critical point(s)

find second derivative substitute the x values and if the value of the derivative is >0 for that value of x then it is a point of minima