Answer by periblepsis for Can you choose the gain AND center the q-point in a common emitter?

voltage gain

If I wanted, say, a gain of 20 or 50 would I have to settle for aq-point that wasn't centered between supply voltage and \$V_{CEmin}\$?

The quiescent emitter voltage must be more than \$\vert A_v\vert\$ times smaller than the quiescent voltage drop across the collector resistor.

A practical circuit must deal with \$V_{_{\text{BE}({q})}}\$ variations between bipolars (perhaps as much as \$\pm 35\:\text{mV}\$), changes in \$V_{_{\text{BE}({q})}}\$ due to operating temperature variation (perhaps as much as \$\pm 90\:\text{mV}\$), possible variations in \$V_{_\text{CC}}\$(ripple, if any.), while also accommodating variations due to \$v_{_{\text{PK}({in})}}\$. To have much of any hope there, the quiescent emitter resistor voltage drop must be on the order of \$500\:\text{mV}\$ -- more being better.

With a \$V_{_\text{CC}}=24\:\text{V}\$ this limits \$\vert A_v\vert\$ to about 20, or so. But that's pushing things when using junkbox parts. I'd stay at or below 10 for this stage topology and your supply voltage (used as an educational lab exercise only since it is a rather poor topology for anything else.)

maximizing \$v_{_{\text{PK}({out})}}\$

Assume \$\vert A_v\vert=10\$ and \$V_{_{\text{CE}({min})}}=1.5\:\text{V}\$(to keep the bipolar in active mode throughout the entire cycle):

1. \$V_{_{\text{RC}({q})}}=\frac12\cdot\frac{24\:\text{V}-1.5\:\text{V}+26\:\text{mV}}{1+\frac1{10}}\approx 10.239\:\text{V}\$
2. \$V_{_{\text{RE}({q})}}=\frac{10.239\:\text{V}}{10}-26\:\text{mV}\approx 997.9\:\text{mV}\$
3. Set \$R_{_\text{C}}=10\:\text{k}\Omega\$, so \$I_{_{\text{C}({q})}}=1.0239\:\text{mA}\$
4. Assume \$\beta=200\pm 50\$,
5. \$R_{_\text{E}}=\frac{997.9\:\text{mV}}{1.0239\:\text{mA}\,\cdot\,\left(1+\frac1{200\pm 50}\right)}\approx 976\:\Omega\$(nearest E96)
6. Assume \$V_{_{\text{BE}({q})}}=680\:\text{mV}\$,
7. \$V_{_{\text{B}({q})}}=976\:\Omega\cdot 1.0239\:\text{mA}\cdot\left(1+\frac1{200\pm 50}\right)+680\:\text{mV}\approx 1.684\:\text{V}\$
8. Using base biasing ratio \$\eta=10\$,
9. \$R_{_\text{B1}}=\frac{1.684\:\text{V}}{\frac{1}{10}\,\cdot\,1.0239\:\text{mA}}\approx 16.5\:\text{k}\Omega\$(nearest E96)
10. \$R_{_\text{B2}}=\frac{24\:\text{V}-1.684\:\text{V}}{1.0239\:\text{mA}\,\cdot\,\left(\frac1{10}+\frac1{200\pm50}\right)}\approx 205\:\text{k}\Omega\$(nearest E96)

Here's the LTspice run:

Note that different BJT values for \$\beta\$ are used, plus a temperature range going from \$-20^\circ\text{C}\$ to \$+55^\circ\text{C}\$. Holds its behavior reasonably well.

Looking back at the calculations note that \$V_{_\text{C(q)}}\approx 13.7\:\text{V}\$ and not \$12\:\text{V}\$. So maximizing \$v_{_{\text{PK}({out})}}\$ doesn't mean\$V_{_\text{C(q)}}=\frac12V_{_\text{CC}}\$.

But this circuit isn't used much outside of education. For a more practical circuit, see the schematic shown at the bottom of this answer.