= Solution
For a <discrete memoryless channel>, its <channel capacity> in bits per use is $C=\max_{P_X}I(X;Y)$. The <Shannon second coding theorem> identifies this with the supremum of rates achievable by arbitrarily long block codes with error <probability> tending to zero: every rate below $C$ is achievable, and no rate above $C$ can have vanishing error. The channel is used independently at each transmission, and coding rate is measured using base-two logarithms.
Let $q$ be the <probability> of using the second input symbol. The second output symbol occurs with <probability> $q/2$, so $H(Y)=h_2(q/2)$. The first input has deterministic output and the second has one bit of conditional output <Shannon entropy>; hence $H(Y\mid X)=q$ and
$$
I(X;Y)=h_2(q/2)-q.
$$
For $0<q<1$, its derivative is $\frac12\log_2[(2-q)/q]-1$, which vanishes at $q=2/5$. <Strict concavity> of <binary entropy> makes this the unique global maximum. Substituting gives
$$
\boxed{C=h_2(1/5)-2/5=\log_2 5-2.}
$$
Thus Shannon's theorem supplies operational coding rates arbitrarily close to this value from below, and rules out reliable transmission above it.
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