Minimize height, above a hole, needed for getting a rectangle out of the hole. I am looking for a solution to a particular task of manipulating a rectangle out of a hole. I will be happy for any suggestions. I wasn't able to find any simplifications (e.g.: transferring the problem to a problem of manipulating a rod), which would lead to an equivalent optimum.
Problem: Minimize height, above a hole, needed for getting a rectangle out of the hole.
Illustration picture:

Having rectangle with known sides
$$
w, h
$$
and having a hole with known width and height:
$$
a, b
$$
Minimize height, above the hole, needed for pulling the rectangle out of the hole
$$
min(x)
$$
EDIT
Two answers (one of them was edited, I will consult it later) suggested placing bottom left/right corner of a rectangle to upper left/right point of the hole and rotating rectangle as much as possible to the right/left. I think this is correct for a "short" rectangle.
I suggest following for a "long" rectangle:
When I take "long" rectangle, I think the height needed (x) is what the picture below shows:

 A: Hint.
A figure worth a thousand words.

NOTE
Attached a MATHEMATICA script which calculates the maximum height clearance needed to free the frame
parms1 = {w -> 1, a -> 1.2, h -> 1.5, b -> 1.9};
parms2 = {w -> 1, a -> 1.6, h -> 1.7, b -> 2.1};
frame = {{-w/2, -h/2}, {w/2, -h/2}, {w/2, h/2}, {-w/2, h/2}};
s = (a - w)/2;
r = (b - h)/2;
g = {gx, gy};
R = RotationMatrix[theta];
ftg = Table[R.frame[[k]] + g, {k, 1, 4}];
lw[lambda_] :=  lambda {-w/2 - s, -h/2 - r} + (1 - lambda) {-w/2 - s, h/2 + r}
rw[lambda_] :=  lambda ftg[[2]] + (1 - lambda) ftg[[3]]
restr1 = (ftg[[1]] - lw[lambda]).(ftg[[1]] - lw[lambda]);
restr2 = ({w/2 + s, h/2 + r} - rw[mu]).({w/2 + s, h/2 + r} - rw[mu]);
restr3 = 0 < lambda < 1;
restr4 = 0 < mu < 1;

obj = ftg[[4]].{0, 1} /. parms1;
restrs = Flatten[{restr1 == 0, restr2 == 0, restr3, restr4}] /. parms1;
sol = NMaximize[Join[{obj}, restrs], {theta, gx, gy, lambda, mu}]
ftg0 = Join[ftg, {ftg[[1]]}] /. parms1 /. sol[[2]];
grfr = ListLinePlot[ftg0];
lw0 = lw[u] /. parms1;
glw = ParametricPlot[lw0, {u, 0, 1}];
grw = ParametricPlot[{w/2 + s, -h/2 - r} u + (1 - u) {w/2 + s, h/2 + r} /. parms1, {u, 0, 1}]; 
Show[grfr, glw, grw, PlotRange -> All, AspectRatio -> 1.8]
x = obj - b/2 /. parms1 /. sol[[2]]

Follows two plots relative to parms1 and parms2 respectively


With $p_1 = (-w/2,-h/2), p_2 = (w/2,-h/2), p_3 = (w/2,h/2), p_4 = (-w/2,h/2)$ we form a frame $F_0 = \{p_1,p_2,p_3,p_4\}$ This frame rotated by $R(\theta)$ and translated by $g = (g_x,g_y)$ is now $F(\theta,g_x,g_y) = R(\theta)\cdot F_0 + g$. The restrictions are $p_1(\theta,g)\in l_w(\lambda),\ 0\le \lambda\le 1$ and $p_0=(a/2,b/2)\in l_r(\mu),\ \ 0\le \mu\le 1$. Here $l_r(\lambda)$ is the left hole wall and $l_r(\mu)$ is the rectangle side with extremum points $p_2(\theta,g)$ and $p_3(\theta,g)$. The clearance is given by $x = p_4(\theta,g)\cdot(0,1)-b/2$
A: Hint: Here is what the diagram should look like

Notice that the left most point of the rectangle is on the line that is with the floor?
Could you take it from there using similar triangles?
Note that there is a gap, because if $a=w$, then the height needed to pull it out would just be $h$, so $a>w$ in this case to solve the not-so-straightforward case.
