La ley de hardy-weinberg- Usar multiplicadores deLaGrange para hacer máximo el valor deP(p, q, r) = 2pq +2pr + 2qr sujeto a p + q + r = 1.
Respuestas
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(1) Hallemos la función de Lagrange
![F(p,q,r,\lambda)=P-\lambda(p+q+r-1)\\ \\
F(p,q,r,\lambda)=(2pq +2pr + 2qr)-\lambda(p+q+r-1)\\ \\
(2)~\texttt{Puntos cr\'iticos de }F\\ \\
F_p=2(q+r)-\lambda=0\\
F_q=2(p+r)-\lambda=0\\
F_r=2(p+q)-\lambda=0\\ \\
\texttt{Sumando: }4(p+q+r)-3\lambda=0\to 4-3\lambda=0\to \lambda =\dfrac{4}{3}\\ \\ \\
(3)~\texttt{De (2) tenemos: }p=q=r=\dfrac{\lambda}{4}=\dfrac{1}{3}\\ \\
\texttt{As\'i }(p,q,r)=\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\texttt{ es un punto cr\'itico} F(p,q,r,\lambda)=P-\lambda(p+q+r-1)\\ \\
F(p,q,r,\lambda)=(2pq +2pr + 2qr)-\lambda(p+q+r-1)\\ \\
(2)~\texttt{Puntos cr\'iticos de }F\\ \\
F_p=2(q+r)-\lambda=0\\
F_q=2(p+r)-\lambda=0\\
F_r=2(p+q)-\lambda=0\\ \\
\texttt{Sumando: }4(p+q+r)-3\lambda=0\to 4-3\lambda=0\to \lambda =\dfrac{4}{3}\\ \\ \\
(3)~\texttt{De (2) tenemos: }p=q=r=\dfrac{\lambda}{4}=\dfrac{1}{3}\\ \\
\texttt{As\'i }(p,q,r)=\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\texttt{ es un punto cr\'itico}](https://tex.z-dn.net/?f=F%28p%2Cq%2Cr%2C%5Clambda%29%3DP-%5Clambda%28p%2Bq%2Br-1%29%5C%5C+%5C%5C%0AF%28p%2Cq%2Cr%2C%5Clambda%29%3D%282pq+%2B2pr+%2B+2qr%29-%5Clambda%28p%2Bq%2Br-1%29%5C%5C+%5C%5C%0A%282%29%7E%5Ctexttt%7BPuntos+cr%5C%27iticos+de+%7DF%5C%5C+%5C%5C%0AF_p%3D2%28q%2Br%29-%5Clambda%3D0%5C%5C%0AF_q%3D2%28p%2Br%29-%5Clambda%3D0%5C%5C%0AF_r%3D2%28p%2Bq%29-%5Clambda%3D0%5C%5C+%5C%5C%0A%5Ctexttt%7BSumando%3A+%7D4%28p%2Bq%2Br%29-3%5Clambda%3D0%5Cto+4-3%5Clambda%3D0%5Cto+%5Clambda+%3D%5Cdfrac%7B4%7D%7B3%7D%5C%5C+%5C%5C+%5C%5C%0A%283%29%7E%5Ctexttt%7BDe+%282%29+tenemos%3A+%7Dp%3Dq%3Dr%3D%5Cdfrac%7B%5Clambda%7D%7B4%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5C%5C+%5C%5C%0A%5Ctexttt%7BAs%5C%27i+%7D%28p%2Cq%2Cr%29%3D%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%2C%5Cdfrac%7B1%7D%7B3%7D%2C%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%5Ctexttt%7B+es+un+punto+cr%5C%27itico%7D)
(4) Ahora veamos si ese punto crítico o estacionario es extremo de la función P (criterio de Sylvester)
(4.1) Hallemos las primeras derivadas de P
![P_p=2(q+r)=\dfrac{4}{3}\\ \\
P_q=2(p+r)=\dfrac{4}{3}\\ \\
P_r=2(p+q)=\dfrac{4}{3}\\ \\ P_p=2(q+r)=\dfrac{4}{3}\\ \\
P_q=2(p+r)=\dfrac{4}{3}\\ \\
P_r=2(p+q)=\dfrac{4}{3}\\ \\](https://tex.z-dn.net/?f=P_p%3D2%28q%2Br%29%3D%5Cdfrac%7B4%7D%7B3%7D%5C%5C+%5C%5C%0AP_q%3D2%28p%2Br%29%3D%5Cdfrac%7B4%7D%7B3%7D%5C%5C+%5C%5C%0AP_r%3D2%28p%2Bq%29%3D%5Cdfrac%7B4%7D%7B3%7D%5C%5C+%5C%5C)
(4.2) Las segundas derivadas de P
![P_{pp}=P_{qq}=P_{rr}=0\\ \\
P_{pq}=P_{pr}=P_{qr}=2 P_{pp}=P_{qq}=P_{rr}=0\\ \\
P_{pq}=P_{pr}=P_{qr}=2](https://tex.z-dn.net/?f=P_%7Bpp%7D%3DP_%7Bqq%7D%3DP_%7Brr%7D%3D0%5C%5C+%5C%5C%0AP_%7Bpq%7D%3DP_%7Bpr%7D%3DP_%7Bqr%7D%3D2)
(4.3) Matriz Hessiana
![\texttt{Sea }g(p,q,r)=p+q+r-1\\ \\
H=\left(\begin{matrix}
0&-g_p&-g_q&-g_r\\
-g_p&F_{pp}&F_{pq}&F_{pr}\\
-g_q&F_{qp}&F_{qq}&F_{qr}\\
-g_r&F_{rp}&F_{rq}&F_{rr}
\end{matrix}\right)=\left(\begin{matrix}
0&-1&-1&-1\\
-1&0&2&2\\
-1&2&0&2\\
-1&2&2&0
\end{matrix}\right)\\ \\ \\
H_3=\left(\begin{matrix}
0&-1&-1\\
-1&0&2\\
-1&2&0
\end{matrix}\right)\to \det(H_1)=4\ \textgreater \ 0\\ \\ \\
H_4=H\to\det{H_4}=-12\ \textless \ 0
\texttt{Sea }g(p,q,r)=p+q+r-1\\ \\
H=\left(\begin{matrix}
0&-g_p&-g_q&-g_r\\
-g_p&F_{pp}&F_{pq}&F_{pr}\\
-g_q&F_{qp}&F_{qq}&F_{qr}\\
-g_r&F_{rp}&F_{rq}&F_{rr}
\end{matrix}\right)=\left(\begin{matrix}
0&-1&-1&-1\\
-1&0&2&2\\
-1&2&0&2\\
-1&2&2&0
\end{matrix}\right)\\ \\ \\
H_3=\left(\begin{matrix}
0&-1&-1\\
-1&0&2\\
-1&2&0
\end{matrix}\right)\to \det(H_1)=4\ \textgreater \ 0\\ \\ \\
H_4=H\to\det{H_4}=-12\ \textless \ 0](https://tex.z-dn.net/?f=%5Ctexttt%7BSea+%7Dg%28p%2Cq%2Cr%29%3Dp%2Bq%2Br-1%5C%5C+%5C%5C%0AH%3D%5Cleft%28%5Cbegin%7Bmatrix%7D%0A0%26amp%3B-g_p%26amp%3B-g_q%26amp%3B-g_r%5C%5C%0A-g_p%26amp%3BF_%7Bpp%7D%26amp%3BF_%7Bpq%7D%26amp%3BF_%7Bpr%7D%5C%5C%0A-g_q%26amp%3BF_%7Bqp%7D%26amp%3BF_%7Bqq%7D%26amp%3BF_%7Bqr%7D%5C%5C%0A-g_r%26amp%3BF_%7Brp%7D%26amp%3BF_%7Brq%7D%26amp%3BF_%7Brr%7D%0A%5Cend%7Bmatrix%7D%5Cright%29%3D%5Cleft%28%5Cbegin%7Bmatrix%7D%0A0%26amp%3B-1%26amp%3B-1%26amp%3B-1%5C%5C%0A-1%26amp%3B0%26amp%3B2%26amp%3B2%5C%5C%0A-1%26amp%3B2%26amp%3B0%26amp%3B2%5C%5C%0A-1%26amp%3B2%26amp%3B2%26amp%3B0%0A%5Cend%7Bmatrix%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0AH_3%3D%5Cleft%28%5Cbegin%7Bmatrix%7D%0A0%26amp%3B-1%26amp%3B-1%5C%5C%0A-1%26amp%3B0%26amp%3B2%5C%5C%0A-1%26amp%3B2%26amp%3B0%0A%5Cend%7Bmatrix%7D%5Cright%29%5Cto+%5Cdet%28H_1%29%3D4%5C+%5Ctextgreater+%5C+0%5C%5C+%5C%5C+%5C%5C%0AH_4%3DH%5Cto%5Cdet%7BH_4%7D%3D-12%5C+%5Ctextless+%5C+0%0A%0A)
![\texttt{Por ende }(p,r,s)=\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\texttt{ es un punto de m\'aximo}
\texttt{Por ende }(p,r,s)=\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\texttt{ es un punto de m\'aximo}](https://tex.z-dn.net/?f=%5Ctexttt%7BPor+ende+%7D%28p%2Cr%2Cs%29%3D%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%2C%5Cdfrac%7B1%7D%7B3%7D%2C%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%5Ctexttt%7B+es+un+punto+de+m%5C%27aximo%7D%0A)
Referencia:
https://es.wikipedia.org/wiki/Multiplicadores_de_Lagrange
(Véase la parte de el caso n-dimensional)
(4) Ahora veamos si ese punto crítico o estacionario es extremo de la función P (criterio de Sylvester)
(4.1) Hallemos las primeras derivadas de P
(4.2) Las segundas derivadas de P
(4.3) Matriz Hessiana
Referencia:
https://es.wikipedia.org/wiki/Multiplicadores_de_Lagrange
(Véase la parte de el caso n-dimensional)
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