Question

Maximize and minimize quantities given an expression with two variables Question Find the difference between the maximum and minimum of the quantity p2q250, where p and q are two nonnegative numbers such that p+q=10. (Enter your answer as a fraction.)

Answer 1

Answer:

The difference between the maximum and minimum is

$\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}$

Step-by-step explanation:

Since p = 10-q, we can replace p in the expression and we get a single-variable function

$f(q)=(10-q)^2q^{250}$

Taking the derivative with respect to q and using the rule for the derivative of a product

$f'(q)=-2(10-q)q^{250}+250(10-q)^2q^{249}$

Critical point (where f'(q)=0)

Assuming q≠ 0 and  q≠ 10

$f'(q)=0\Rightarrow -2(10-q)q^{250}+250(10-q)^2q^{249}=0\Rightarrow\\\\\Rightarrow 250(10-q)=2q\Rightarrow q=\displaystyle\frac{625}{63}$

To check this is maximum, we take the second derivative

$f''(q)=63252q^{250}-1255000q^{249}+6225000q^{248}$

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

$p^2q^{250}$

is

$\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}$

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

$\left(\displaystyle\frac{5}{63}\right)^2\left(\displaystyle\frac{625}{63}\right)^{250}$