Heat transfer direction in fins

I’m unsure of what exactly is changing the heat transfer direction in the following triangular fin:
$$
q_{x} = -kA(x)frac{mathrm{d}T(x)}{mathrm{d}x}
tag{1}
$$
$$
begin{align}
q_{x+mathrm{d}x} &= q_{x}+frac{partial q_{x}}{partial x}mathrm{d}x
&=-kA(x)frac{mathrm{d}T(x)}{mathrm{d}x} – kfrac{partial}{partial x}[A(x)frac{mathrm{d}T(x)}{mathrm{d}x}]mathrm{d}x
tag{2}
end{align}
$$
$$
mathrm{d}q_{text{conv}} = h(x)mathrm{d}S(x)[T(x) – T_{∞}]
tag{3}
$$
where all the symbols are reported below:

$q_x, q_{text{conv}}$ heat transfer rate $[mathrm{W}]$.

$h(x)$ convection heat transfer coefficient $[mathrm{W,m^{-2},K^{-1}}]$.

$k$ thermal conductivity $[mathrm{W,m^{-1},K^{-1}}]$.

$T(x)$ temperature $[mathrm{K}]$.

$T_b=T(x=0)$ base surface temperature $[mathrm{K}]$.

$T_infty$ bulk temperature $[mathrm{K}]$.

$A(x)$ cross-sectional area $[mathrm{m}^2]$.

$A_0=A(x=0)$ base area $[mathrm{m}^2]$.

$mathrm{d}S(x)$ surface area of the differential element $[mathrm{m}^2]$.

$L$ total fin length $[mathrm{m}]$.

Writing down the indefinite energy balance and assuming that $T_b>T(x)>T_{infty}$:
$$
0=q_{text{in}}-q_{text{out}}= q_{x} – (mathrm{d}q_{text{conv}} + q_{x+mathrm{d}x})
tag{4}
$$
Substitute $(1)$, $(2)$ and $(3)$ in $(4)$:
$$
begin{align}
q_{x} &= mathrm{d}q_{text{conv}} + q_{x+mathrm{d}x}[5pt]
-kA(x)frac{mathrm{d}T(x)}{mathrm{d}x} &= h(x)mathrm{d}S(x)[T(x) – T_{∞}] -kA(x)frac{mathrm{d}T(x)}{mathrm{d}x} – kfrac{partial}{partial x}[A(x)frac{mathrm{d}T(x)}{mathrm{d}x}]mathrm{d}x
0 &= h(x)mathrm{d}S(x)[T(x) – T_{∞}] – kfrac{partial}{partial x}[A(x)frac{mathrm{d}T(x)}{mathrm{d}x}]mathrm{d}x
end{align}
tag{5}
$$
introducing the adimensional lenght $X$, we obtain for every function $f(x)=f(LX)=f(X)$, including any differential relation:
$$
X overset{Delta}{=} x/Lto x=LXquadtext{therefore}quadmathrm{d}X=mathrm{d}x/Ltomathrm{d}x=Lmathrm{d}X
$$
So, four new functions are defined, where $bar h$ is the averaged convection heat transfer coefficient on $X$. In the following list, all of the uppercase roman new functions are adimensional:
$$
theta(X) overset{Delta}{=} frac{T(X) – T_{∞}}{T_{b} – T_{∞}}quad K(X) overset{Delta}{=} frac{A(X)}{A_{0}} quad p(X) overset{Delta}{=} frac{mathrm{d}S(X)}{mathrm{d}X}
$$
$$
W(X) overset{Delta}{=} frac{h(X)}{p_{0}bar h}frac{mathrm{d}S(X)}{mathrm{d}X} = frac{h(X)}{bar h}frac{p(X)}{p(0)}
$$
Then two constants are introduced (uppercase is adimensional):
$$
m = frac{bar h p(0)}{kA_0}quad M = mL
$$
Substituting into $(5)$ and rearranging we obtain the adimensional differential relation:
$$
begin{align}
h(X)frac{mathrm{d}S(X)}{mathrm{d}x}[T(X)-T_infty] &= kfrac{partial}{partial x}[A(X)frac{mathrm{d}T(X)}{mathrm{d}x}][5pt]
h(X)frac{mathrm{d}S(X)}{mathrm{d}X}theta(X) &= kfrac{mathrm{d}}{mathrm{d}X}[A(X)frac{mathrm{d}theta(X)}{Lmathrm{d}X}][5pt]
h(X)p(X)theta(X) &= frac{kA_0}{L}frac{mathrm{d}}{mathrm{d}X}[K(X)frac{mathrm{d}theta(X)}{mathrm{d}X}][5pt]
W(X)theta(X) &= frac{1}{mL}frac{mathrm{d}}{mathrm{d}X}[K(X)frac{mathrm{d}theta(X)}{mathrm{d}X}][5pt]
MW(X)theta(X) &= frac{mathrm{d}}{mathrm{d}X}[K(X)frac{mathrm{d}theta(X)}{mathrm{d}X}]
end{align}
tag{6}
$$
Now, using a rectangular profile, it’s clear that $A(X)=A_0$ and $K(X)=1$, assuming that $W(X)=1$ we obtain:

                   
$$
Mtheta(X) = frac{mathrm{d}}{mathrm{d}X}[frac{mathrm{d}theta(X)}{mathrm{d}X}]
$$

Using a triangular profile, we have $A(X)neq A_0$ and $K(X)simeq X$, assuming that $W(X)=1$ we obtain:

                              
$$
Mtheta(X) = frac{mathrm{d}}{mathrm{d}X}[Xfrac{mathrm{d}theta(X)}{mathrm{d}X}]
$$

I did the energy balance thinking of the heat going from left to right like it’s shown on the rectangular fin, but on the triangular fin it goes on the opposite direction.

Which is fine for what I want to do, but why exactly does this happen?

Due to the equations involved the heat will only move from a bigger to an equal or smaller area?

Which adjustments would I have to make to change the heat transfer direction and make a fin like this:

                                   

Would such fin make any sense?