Chapter 14 - Section 14.7 - Maximum and Minimum Values - 14.7 Exercise - Page 969: 58

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, $P=\dfrac{2}{3}$

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