# Thread: Maximize and Minimize Area

1. ## Maximize and Minimize Area

Please help!!! I should know how to do this but it's just not happening for me.

If you have 12 meters of wire that has to be cut into 2 pieces. 1 piece is to be bent into the shape of a circle and the other into the shape of a square. How should the wire be cut so as to: a) maximize the total area enclosed by the shapes. b) Minimize the total area enclosed by the shapes.

Thanks for your help.

2. Hi
Originally Posted by Gigabyte
If you have 12 meters of the wire that has to be cut into 2 pieces. 1 piece is to be bent into the shape of a circle and the other into the shape of a square.
Let's call $L_{circ}$ the length of wire bent into the shape of a circle. $L_{circ}$ is the perimeter of the circle and it is linked to its radius $R$ by $L_{circ}=2\pi R \implies R=\frac{L_{circ}}{2\pi}$. Thus the area of the circle is $A_{circ}=\pi R^2=\pi \frac{L_{circ}^2}{4\pi ^2}=\frac{L_{circ}^2}{4\pi}$

Let's do the same thing for the square : let's call $L_{square}$ the length of wire bent into the shape of a square. $L_{square}$ is the perimeter of the square thus the length of one of its side is $\frac{L_{square}}{4}$ and its area is $\left(\frac{L_{square}}{4}\right)^2= \frac{L_{square}^2}{4^2}$

As the length of wire is $12\mathrm{m}$, $L_{square}+L_{circle}=12 \implies L_{circle}=12-L_{square}$. The area of the two shapes can be given in function of $L_{square}$ :

$A=\frac{L_{square}^2}{4^2}+\frac{L_{circ}^2}{4\pi} =\frac{L_{square}^2}{4^2}+\frac{(12-L_{square})^2}{4\pi}$

The problem boils down to finding the maximum and the minimum of the function $A(x)=\frac{x^2}{4^2}+\frac{(12-x)^2}{4\pi}$ for $x\in[0,\,12]$. (because $0 \leq L_{square} \leq \L_{wire}=12$ )

Can you take it from here ?

3. thanks mate.

But wouldn't the area of the square be
(Lsquare/4)^2 not (Lsquare/4)^3?

And then the differentiation of A i couldn't do. Well i've tried and i get x=6.53.

4. Originally Posted by Gigabyte
thanks mate.

But wouldn't the area of the square be
(Lsquare/4)^2 not (Lsquare/4)^3?
Yes

And then the differentiation of A i couldn't do. Well i've tried and i get x=6.53.
$\frac{\mathrm{d}A}{\mathrm{d}x}=2\frac{x}{4^2}-\frac{2(12-x)}{4\pi}=x\left(\frac{1}{8}+\frac{1}{2\pi}\right)-\frac{6}{\pi}=x\cdot\frac{\pi+4}{8\pi}-\frac{6}{\pi}$

This derivative equals 0 if $x=\frac{48}{\pi+4}$, it is negative for $0\leq x \leq \frac{48}{\pi+4}$ and positive if $\frac{48}{\pi+4}\leq x \leq 12$ hence $A$ is increasing on $\left[\frac{48}{\pi+4},\,12 \right]$ and decreasing on $\left[0,\,\frac{48}{\pi+4} \right]$. It can be deduced from this that the area is minimized at $x_m=\frac{48}{\pi+4}$ and maximized at $x_M=0$ or $x_M=12$. To choose between this two values, one can compare $A(0)$ and $A(12)$ : $x_M=0$ if $A(0)>A(12)$ and $x_M=12$ otherwise.