From QED

To begin, we need to realize that all species will be fully charged, since our temperatures are in the 10s of KeV, far above any ionization potential for light atoms. Boron has a fully charged state of 5. If you don't know this, it can be deduced from the reaction equation (1 proton plus protons from boron equals 6 protons. so boron has 5 protons). We also know that

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \mathrm{P}_\mathrm{fusion} \propto n_\mathrm{H}n_\mathrm{B},

as well as

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \beta \propto n_\mathrm{e} + n_\mathrm{H} + n_\mathrm{B}.

From charge conservation,

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): n_\mathrm{e} = n_\mathrm{H} + 5n_\mathrm{B},

so

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \beta \propto 2( n_\mathrm{H} + 3n_\mathrm{B}).

We can use this to eliminate one of the densities from our power equation,

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \mathrm{P}_\mathrm{fusion} \propto \left(\frac{\beta}{2} - 3n_\mathrm{B}\right)n_\mathrm{B}.

We can now maximize the fusion power by taking a derivative, finding

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): 0 = \frac{\beta}{2} - 6n_\mathrm{B} = n_\mathrm{H} - 3n_\mathrm{B}.

So

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): n_\mathrm{H} = 3n_\mathrm{B}.

To complete the problem, we use the charge conservation equation.

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \frac{n_\mathrm{H}}{n_\mathrm{e}} =\frac{3}{16}

Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \frac{n_\mathrm{B}}{n_\mathrm{e}} =\frac{1}{8}