Three alleles (alternative versions of a gene) A, B, and O determine the four blood types A (AA or AO), B (BB or BO), O (OO), and AB. The Hardy-Weinbe

Maximum value of $P=\frac{2}{3}$
1) Concept:
Use the concept of absolute maximum and minimum.
To determine a continuous function's exact maximum or minimum value on the closed bonded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D
3. The absolute maximum and lowest values of steps 1 and 2 are represented by their largest and smallest values, respectively.
2) Given:
Hardy-Weinberg Law:
The proportion of individuals in a population who carry two different allers is
$P=2pq+2pr+2rq$
Where p, q and r be A, B and O blood group populations.
$p+qr=1$
3) Calculation:
Consider Hardy-Weinberg Law,
The proportion of individuals in a population who carry two different alleles is
$P=2pq+2pr+2rq$
Where p, q and r be the A, B and O blood group populations.
By using fact, $p+q+r=1$
write r in terms of p and q
$r=1-p-q$
Substitute this value of r in $P=2pq+2pr+2rq$, then it becomes
$P=P(p,q)=2(1-p-q)q+2(1-p-q)p+2pq$
$P(p,q)=2p-2p^2+2q-2q^2-2pq$
As p, q, r is the proportion of species it ranges from 0 to 1.
Therefore, $p\ge 0,q\ge 0,1-p-q\ge 0$ this implies
$p+q\le 1$
Therefore, domain of P is
$D=(p,q)0\le p\le 1,q\le 1-p$ which is closed set
bounded by lines $p=0,q=0andp+q=1.$
Now to find critical points, consider equation
$P(p,q)=2p-2p^2+2q-2q^2-2pq$
Differentiating P partially with respect to p,
$P_p(p,q)=2-4p-2q$
Differentiating P partially with respect to q,
$P_q(p,q)=2-4q-2p$
Setting partial derivatives equal to 0, obtain the equations,
$2-4p-2q=0\text{ and }2-4q-2p=0$
Therefore,
$2p+q=1\text{ and }p+2q=1$
Solving these system of equations,
$p+2(1-2p)=1$
$p=\frac{1}{3}$
Substitute $p=\frac{1}{3}\text{ in }2p+q=1$
$q=1-2(\frac{1}{3})=\frac{1}{3}$
Thus the critical point is $(\frac{1}{3},\frac{1}{3})$
Therefore,
$p(\frac{1}{3},\frac{1}{3})=2(\frac{1}{3})-2(\frac{1}{3}{)}^2+2(\frac{1}{3})-2(\frac{1}{3}{)}^2-2(\frac{1}{3})(\frac{1}{3})$
$=\frac{4}{1}-\frac{6}{9}=\frac{12-6}{9}=\frac{6}{9}=\frac{2}{3}$
$p(\frac{1}{3},\frac{1}{3})=\frac{2}{3}$
Now find values of P on the boundary of D consisting three lines.
Along the line $p=0$, q ranges between 0 and 1 that is
$0\le q\le 1.$
Therefore, $P(0,q)=2q-2q^2$
Which represents downward parabola with maximum value at vertex $(0,\frac{1}{2})$
$P(\frac{1}{2},0)=2(\frac{1}{2})-2(\frac{1}{2}{)}^2+2(0)-2(0{)}^2-2(0)(\frac{1}{2})=\frac{1}{2}$
Therefore,
$P(0,\frac{1}{2})=\frac{1}{2}$
Similarly, along the line $q=0$,
p ranges between 0 and 1 that is $0\le p\le 1.$
Therefore, $P(p,0)=2p-2p^2$
Which represents downward parabola with maximum value at vertex $(\frac{1}{2},0)$