How to prove that the following function is maximized when $p = q$?

Question

Given the following function:

\begin{equation} f(q,p) = q\log(2p) + (1 - q)\log(2 - 2p) \end{equation}

Given some $q \in [0,1]$ we need to prove that $f(q, p)$ is maximized when $p = q$. How to prove this?

Answer 1

Since $q\in[0,1]$ is given we have to study the function $$g_q(p):=q\log p+(1-q)\log(1-p)+\log 2$$ on the interval $0\leq p\leq1$. One has $g_0(p)=\log (1-p)$, which is maximal at $p=0$, as claimed, and similarly $g_1(p)=\log p$ is maximal at $p=1$.

When $0<q<1$ then both $\lim_{p\to0+} g_q(p)=-\infty$, $\lim_{p\to1-} g_q(p)=-\infty$. Hence the function $g_q$ assumes its maximum at an interior point $p_*\in\>]0,1[\>$, which will be brought to the fore by letting $$g_q'(p)={q\over p}-{1-q\over 1-p}={q-p\over p(1-p)}=0\ .$$It follows that $p_*=q$, as claimed.

Answer 2

Take the gradient $$ \nabla f(p,q)=\left(\log(2p)-\log(2(1-p)),\frac qp-\frac{(1-q)}{(1-p)}\right) $$ and putting this equal to $(0,0)$; thus you get $$ \log(2p)=\log(2(1-p))\\ \frac qp=\frac{(1-q)}{(1-p)} $$ i.e. \begin{align*} p&=1-p\;\;\Longrightarrow p=1/2\\ q/(1/2)&=(1-q)/(1/2)\;\;\Longrightarrow q=1/2\;\;. \end{align*}

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Let's instead consider the following problem: fixed $q\in[0,1]$, maximize $p\mapsto f(p,q)$ on $[0,1]$.

Let's thus consider the $p$-derivative: $$ \partial_pf(p,q)=\log(2p)-\log[2(1-p)] $$ which is zero iff $p=1-p$, i.e. $p=1/2$