# Question #30b71

Dec 2, 2016

The area is maximum if all the wire is used for the square.
it is minimum when the side of the square is $\frac{20 \sqrt{3}}{9 + 4 \sqrt{3}}$

#### Explanation:

Name $l$ the length of the piece that is used for the square.
The piece that is used for the triangle is then $10 - l$.

The area of the square is:
${A}_{1} = {\left(\frac{l}{4}\right)}^{2} = {l}^{2} / 16$

The area of the triangle is:
${A}_{2} = \frac{1}{2} \frac{10 - l}{3} \cdot \frac{10 - l}{3} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{36} {\left(10 - l\right)}^{2}$

The total area is:

$A \left(l\right) = {A}_{1} + {A}_{2} = {l}^{2} / 16 + \frac{\sqrt{3}}{36} {\left(10 - l\right)}^{2}$

$\frac{\mathrm{dA}}{\mathrm{dl}} = \frac{l}{8} - \frac{\sqrt{3}}{18} \left(10 - l\right)$

$\frac{{d}^{2} A}{\mathrm{dl}} ^ 2 = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0$

Equate the derivative of $A \left(l\right)$ to zero to find critical points:

$\frac{l}{8} - \frac{\sqrt{3}}{18} \left(10 - l\right) = 0$

$\frac{l}{8} - 5 \frac{\sqrt{3}}{9} + \frac{\sqrt{3}}{18} l = 0$

$l \left(\frac{9}{8} + \frac{\sqrt{3}}{2}\right) = 5 \sqrt{3}$

$l = \frac{80 \sqrt{3}}{9 + 4 \sqrt{3}}$

As the second derivative is always positive, this point is the minimum.

As there are no other stationary points, the maximum will be at one the extreme points of the interval, i.e. either for $l = 0$ or $l = 10$.

$A \left(0\right) = \frac{5 \sqrt{3}}{18}$

$A \left(10\right) = \frac{25}{4} > A \left(0\right)$