Maximum value of \(P = \frac{2}{3}\)

1) Concept:

Use the concept of absolute maximum and minimum.

To find the absolute maximum or minimum value of a continuous function f 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 largest value of step 1 and step 2 is absolute maximum value and smallest value being the absolute minimum value.

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.

\(p + q r =1\)

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, \(p \geq 0, q \geq 0, 1 - p - q \geq 0\) this implies

\(p + q \leq 1\)

Therefore, domain of P is

\(D = {(p, q) 0 \leq p \leq 1, q \leq 1 - p}\) which is closed set

bounded by lines \(p = 0, q = 0 and p + q = 1.\)

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,

\(2 - 4p - 2q = 0 and 2 - 4q - 2p = 0\)

Therefore,

\(2p + q = 1 and p + 2q = 1\)

Solving these system of equations,

\(p + 2(1 - 2p) = 1\)

\(p = \frac{1}{3}\)

Substitute \(p = \frac{1}{3} in 2p + q = 1\)

\(q =1 - 2(\frac{1}{3}) = \frac{1}{3}\)

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

\(0 \leq q \leq 1.\)

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 \(0 \leq p \leq 1.\)

Therefore, \(P(p, 0) = 2p - 2p^{2}\)

Which represents downward parabola with maximum value at vertex \((\frac{1}{2}, 0)\)

\(P(\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} - 2(0)(\frac{1}{2}) = \frac{1}{2}\)

Therefore,

\(P(\frac{1}{2}, 0) = \frac{1}{2}\)

Also along line \(p + q = 1\)

where \(0 \leq p \leq 1,\)

\(P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)\)

\(= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \leq p \leq 1\)

\(P(p, 1 - p) = 2p - 2p^{2}\)

Which is downward parabola whose vertex is at \((\frac{1}{2}, \frac{1}{2})\)

Therefore, maximum value is

\(P(\frac{1}{2}, \frac{1}{2}) = 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} = \frac{1}{2}\)

Therefore, comparing all values at boundary with value of P at the critical point

The absolute maximum value of P(p, q) on D is \(\frac{2}{3}.\)

Therefore, the maximum value of P is \(\frac{2}{3}.\)

Conclusion:

The maximum value of P is \frac{2}{3}.

1) Concept:

Use the concept of absolute maximum and minimum.

To find the absolute maximum or minimum value of a continuous function f 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 largest value of step 1 and step 2 is absolute maximum value and smallest value being the absolute minimum value.

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.

\(p + q r =1\)

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, \(p \geq 0, q \geq 0, 1 - p - q \geq 0\) this implies

\(p + q \leq 1\)

Therefore, domain of P is

\(D = {(p, q) 0 \leq p \leq 1, q \leq 1 - p}\) which is closed set

bounded by lines \(p = 0, q = 0 and p + q = 1.\)

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,

\(2 - 4p - 2q = 0 and 2 - 4q - 2p = 0\)

Therefore,

\(2p + q = 1 and p + 2q = 1\)

Solving these system of equations,

\(p + 2(1 - 2p) = 1\)

\(p = \frac{1}{3}\)

Substitute \(p = \frac{1}{3} in 2p + q = 1\)

\(q =1 - 2(\frac{1}{3}) = \frac{1}{3}\)

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

\(0 \leq q \leq 1.\)

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 \(0 \leq p \leq 1.\)

Therefore, \(P(p, 0) = 2p - 2p^{2}\)

Which represents downward parabola with maximum value at vertex \((\frac{1}{2}, 0)\)

\(P(\frac{1}{2}, 0) = 2(0) - 2(0)^{2} + 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} - 2(0)(\frac{1}{2}) = \frac{1}{2}\)

Therefore,

\(P(\frac{1}{2}, 0) = \frac{1}{2}\)

Also along line \(p + q = 1\)

where \(0 \leq p \leq 1,\)

\(P(p, q) = P(p, 1 - p) = 2p - 2p^{2} + 2(1 - p) - 2(1 - p)^{2} - 2p(1 - p)\)

\(= 2p - 2p^{2} + 2 - 2p - 2 + 4p - 2p^{2} - 2p + 2p^{2}, 0 \leq p \leq 1\)

\(P(p, 1 - p) = 2p - 2p^{2}\)

Which is downward parabola whose vertex is at \((\frac{1}{2}, \frac{1}{2})\)

Therefore, maximum value is

\(P(\frac{1}{2}, \frac{1}{2}) = 2(\frac{1}{2}) - 2(\frac{1}{2})^{2} = \frac{1}{2}\)

Therefore, comparing all values at boundary with value of P at the critical point

The absolute maximum value of P(p, q) on D is \(\frac{2}{3}.\)

Therefore, the maximum value of P is \(\frac{2}{3}.\)

Conclusion:

The maximum value of P is \frac{2}{3}.