Optimise and differentiate a triangle on an implicit curve

At $y=0$, $4x^2=36$, so $P=(-3,0)$ and $R=(3,0)$. The base length is $6$ and perpendicular height is $y$, hence $A=\tfrac12(6)y=3y$.

$8x+xy^{\prime}+y+2yy^{\prime}=0$, so $(x+2y)y^{\prime}=-(8x+y)$ and $dy/dx=-(8x+y)/(x+2y)$ wherever $x+2y\ne0$.

$A^{\prime}=3y^{\prime}=0$ gives $y=-8x$. Substitute into the curve: $4x^2-8x^2+64x^2=36$, so $x^2=3/5$. The positive root would make $y<0$, so reject it. Thus $x=-\sqrt{3/5}$ and $y=8\sqrt{3/5}$.

Differentiate $8x+(x+2y)y^{\prime}+y=0$:

$\begin{aligned}&8+(x+2y)y^{\prime\prime}+2y^{\prime}+2(y^{\prime})^2\\ & =0.\end{aligned}$

At the stationary point, $y^{\prime}=0$, $y=-8x$, so $y^{\prime\prime}=-8/(x+2y)=8/(15x)<0$. Since $A^{\prime\prime}=3y^{\prime\prime}<0$, the stationary area is a maximum.

At $x=0$, the upper curve has $y=6$. The differentiated equation gives $y^{\prime}=-1/2$, so $dA/dt=3(dy/dx)(dx/dt)=3(-1/2)(8)=-12$ square units/s. The area decreases at $12$ square units/s.

$8x\dot x+x\dot y+y\dot x+2y\dot y=0$. At $x=0$, $y\dot x+2y\dot y=0$, so $\dot y=-\dot x/2=-4$. Hence $\dot A=3\dot y=-12$ square units/s, again a decrease of $12$.