Answer
a) the maximum are occurs when $x$ = $10$ and all the wire is used for the square
b) the minimum area occurs when $x$ = $\frac{40\sqrt 3}{9+4\sqrt 3}$ $\approx$ $4.35$ $m$
Work Step by Step
Let $x$ be the length of the wire used for the square
The total area is
$A(x)$ = $\left(\frac{x}{4}\right)^{2}+\frac{1}{2}\left(\frac{10-x}{3}\right)\left(\frac{\sqrt 3}{2}\right)\left(\frac{10-x}{3}\right)$ = $\frac{1}{16}x^{2}+\frac{\sqrt 3}{36}(10-x)^{2}$, $0$ $\leq$ $x$ $\leq$ $10$
$A'(x)$ = $\frac{1}{8}x-\frac{\sqrt 3}{18}(10-x)$
$A'(x)$ = $0$
$\frac{1}{8}x-\frac{\sqrt 3}{18}(10-x)$ = $0$
$x$ = $\frac{40\sqrt 3}{9+4\sqrt 3}$
Now
$A(0)$ = $\left(\frac{\sqrt 3}{36}\right)100$ $\approx$ $4.81$
$A(10)$ = $\frac{100}{16}$ = $6.25$
$A\left(\frac{40\sqrt 3}{9+4\sqrt 3}\right)$ $\approx$ $2.72$
a) the maximum are occurs when $x$ = $10$ and all the wire is used for the square
b) the minimum area occurs when $x$ = $\frac{40\sqrt 3}{9+4\sqrt 3}$ $\approx$ $4.35$ $m$
