Answer
Absolute maximum: $\dfrac{2}{3}$
Work Step by Step
Given: $P=2pq+2pr+2rq$
From question: $p+q+r=1$ this gives $r=1-p-q$
Thus,
$P=2pq+2p(1-p-q)+2(1-p-q)q$
or, $P=-2p^2-2q^2-2pq+2p+2q$
Solve for first partial derivative.
Therefore, $2p+q=1$ and $2q+p=1$
$\implies$ $q=1-2p$
After simplification, we get $p=q=r=\dfrac{1}{3}$
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point.
$D=12 \gt 0$ and $f_{pp}=-4 \lt 0$
$D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
Thus, $P=2(\dfrac{1}{3})(\dfrac{1}{3})+2(\dfrac{1}{3})(\dfrac{1}{3})+2(\dfrac{1}{3})(\dfrac{1}{3})$
or, $P=\dfrac{2}{3}$
