The problem consist in find every value of a,b,c,d and H. (D=7d)
Ia already know that for the first piece of the funnel, $A=\pi\left(\frac{7d}{2}\right)^2$, then for the third piece $A=\pi(\frac{d}{2})^2$. As the angle is $45°$ this arrives to the triangular central parts are isoceles and then $b=3d$, and the area as a function of height are given by $A(h)=\pi(h+d/2)^2$ in the whole region of lenght "b".
Finally, when the time is zero the total volume of wine at the funnel are 900mL cuz 750(1.2)=900. Now I'm using Torricelli's Law for this problem, and I arrive to: For the first part: $$\pi\left(\frac{7d}{2}\right)^2\frac{dh}{dt}=-\pi(d/2)^2\sqrt{2gh}\Rightarrow 2\sqrt{h}=\sqrt{2g}t+C_1$$ ($C_1=\sqrt{H}$ at time $t=0$)
Proceding on similar way I get the relation for second and third part of the funnel, but I don't know how continue cuz I can't see the way to use the total volume to find al the values requested. Any hint or solution?
Also, am I correct thinking in take $A(h)$ as a piecewise function realizing that the funnel are irregular?
Hint.
Considering $(S_1,h_1), (S_2,h_2)$ respectively as the top, height and bottom, height surfaces we have:
$$ \cases{ S_1(h_1)\frac{dh_1}{dt}=S_2(h_2)\frac{dh_2}{dt},\ \text{Conservation of mass}\\ \frac 12\delta \left(\frac{dh_2}{dt}\right)^2=(h_1-h_2)\delta g, \ \text{Torricelli} } $$ then
$$ \frac{dh_1}{dt}=\frac{S_2(h_2)}{S_1(h_1)}\sqrt{2g(h_1-h_2)} $$
now considering $h$ generic and $h_2=0, S_2(0) = S_0$ we have
$$ \frac{dh}{dt}=S_0\frac{\sqrt{2g h}}{S_1(h)} $$
This ODE is separable so we have
$$ S_0^{-1}\frac{S(h)}{\sqrt{2g h}}dh = dt $$
and then
$$ S_0^{-1}\int_0^H \frac{S(h)}{\sqrt{2g h}}dh = \Delta t = \Phi(a,b,c,D,d) $$
so now we can calculate
$$ \min\Delta t = \min_{a,b,c,D.d}\Phi(a,b,c,D,d),\ \ \text{s. t.}\ R(a,b,c,D,d) $$
where $R(a,b,c,D,d)$ represents the restriction set.
NOTE
Here $S(h)$ is a convenient piecewise function and
$$ \int_0^H = \int_0^c+\int_c^{c+b}+\int_{c+b}^H = \pi \left(\frac d2\right)^2\sqrt{\frac{2c}{g}}+\int_c^{c+b}\frac{S(h)}{\sqrt{2g h}}dh+\frac{2\pi}{\sqrt{2g}}\left(\frac D2\right)^2\left(\sqrt{a+b+c}-\sqrt{b+c}\right) $$