Minimum Area Enclosed

Relevant wiki: Optimization

Let $r$ be the radius of the circle and $a$ be the side length of the square.

Recall the perimeter of a circle and a square can be expressed as $2\pi r$ and $4a$ , respectively.

Then we can set up the following equation regarding the fence used:

$(2\pi r + 4r) + 4a = 1000$

Changing the subject to $a$ gives $a = 250 - \left( 1 + \dfrac \pi2 \right) r$ .

The combined area:

$A = r^2π + a^2 = r^2π + (250 - (1+0.5π)r)^2$

An easy way to find the minimal area is by differentiating (using the chain rule):

$\frac {dA}{dR} = 2r \pi - 2(250 - (1+0.5π)r)(1+0.5π)$

$2r \pi - 2(250 - (1+0.5 \pi )r)(1+0.5 \piπ) = 0$

$r = \dfrac {500(1+0.5 \pi)}{ 2 \pi+2(1+0.5 \pi)^2} \approx 65.91389 \text{ m}$

Substituting this value back into our formulae regarding a and A, we get:

$a \approx 80.54881 \text{ m}$

and

$A_\text{min} = 20137.20331 \text{ m}^2 = \boxed {20137 \text{ m}^2 \text { (0 d. p.) } }$ .

By the second derivative test , we can show that $\left . \frac {d^2 A}{dR^2} \right |_{r \approx 65.91389}> 0$ , so the extremal point that we have found is indeed the minimum value.

x = length of wire bent into a circle

1000 - x = length of wire bent into a square

$x = πd + 2d$

$d=$ $\frac{x}{π+2}$

$A=$ $\frac{π}{4}$ $($ $\frac{x}{π+12})^2$ $+($ $\frac{1000-x}{4}$ $)^2$

$A=$ $0.029709x^2$ $+$ $\frac{(1000-x)^2}{16}$

$\frac{dA}{dx}$ $=0.059418x$ $-\frac{1000-x}{8}$ $=0$

$x=677.8047471$

$d=131.8277804$

$A=13649.09199+6488.111312=20137.2033$