Answer:
The difference between the maximum and minimum is
[tex]\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}[/tex]
Step-by-step explanation:
Since p = 10-q, we can replace p in the expression and we get a single-variable function
[tex]f(q)=(10-q)^2q^{250}[/tex]
Taking the derivative with respect to q and using the rule for the derivative of a product
[tex]f'(q)=-2(10-q)q^{250}+250(10-q)^2q^{249}[/tex]
Critical point (where f'(q)=0)
Assuming q≠ 0 and q≠ 10
[tex]f'(q)=0\Rightarrow -2(10-q)q^{250}+250(10-q)^2q^{249}=0\Rightarrow\\\\\Rightarrow 250(10-q)=2q\Rightarrow q=\displaystyle\frac{625}{63}[/tex]
To check this is maximum, we take the second derivative
[tex]f''(q)=63252q^{250}-1255000q^{249}+6225000q^{248}[/tex]
and
f''(625/63) < 0
so q=625/63 is a maximum. For this value of q we get p=5/63
The maximum value of
[tex]p^2q^{250}[/tex]
is
[tex]\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}[/tex]
The minimum is 0, which is obtained when q=0 and p=10 or q=10 and p=0
The difference between the maximum and minimum is then
[tex]\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}[/tex]
