Perimeter:
$P = 2y + x$
$0 = 2y' + 1$
$y' = -1/2 = -0.5$
Area:
$A = \frac{1}{2} xh$
From the figure:
$h = \sqrt{y^2 \, - \, 0.25x^2}$
Thus,
$A = \frac{1}{2}x \sqrt{y^2 - 0.25x^2}$
$\dfrac{dA}{dx} = \dfrac{1}{2} \left[ x \left( \dfrac{2y \, y' - 0.5x}{2 \sqrt{y^2 - 0.25x^2}} + \sqrt{y^2 - 0.25x^2}\right) \right] = 0$
$\dfrac{2(xy\,y' - 0.25x^2)}{2\sqrt{y^2 - 0.25x^2}} + \sqrt{y^2 - 0.25x^2} = 0$
multiply both sides of the equation by $\sqrt{y^2 - 0.25x^2}$
$xy\,y' - 0.25x^2 + y^2 - 0.25x^2 = 0$
$xy(-0.5) - 0.5x^2 + y^2 = 0$
$y^2 - 0.5xy - 0.5x^2 = 0$
$2y^2 - xy - x^2 = 0$
Solving for y by quadratic formula: a = 2; b = -x; c = -x2
$y = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$y = \dfrac{-(-x) \pm \sqrt{x^2 - 4(2)(-x^2)}}{2(2)}$
$y = \dfrac{x \pm \sqrt{x^2 + 8x^2}}{4}$
$y = \dfrac{x \pm 3x}{4}$
$y = x \, \text{ and } \, -\frac{1}{2}x$
y = -½ x is absurd, thus use y = x
Therefore
$P = 2x + x = 3x$
$x = y = \frac{1}{3}P$
$h = \sqrt{y^2 \, - \, 0.25x^2}$
$h = \sqrt{(\frac{1}{3}P)^2 - \frac{1}{4}(\frac{1}{3}P)^2}$
$h = \sqrt{\frac{1}{9}P^2 - \frac{1}{36}P^2}$
$h = \sqrt{\frac{1}{12}P^2}$
$h = \frac{1}{2\sqrt{3}}P$
$A = \frac{1}{2}xh$
$A_{max} = \frac{1}{2}(\frac{1}{3}P)(\frac{1}{2\sqrt{3}}P)$
$A_{max} = \frac{1}{12\sqrt{3}}P^2$
$A_{max} = \frac{1}{12\sqrt{3}} \, \cdot \, \frac{\sqrt{3}}{\sqrt{3}}P^2$
$A_{max} = \frac{\sqrt{3}}{36}P^2$ answer
