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

To maximize and minimize the quantity given by p^2q^250 with the constraint p+q=10, express q in terms of p, derive a single-variable function, optimize using differentiation, find critical points, and evaluate at the endpoints of the domain [0, 10]. The difference between the maximum and minimum values would be the answer.

Step-by-step explanation:

The student asks how to maximize and minimize the quantity p2q250, given the constraint that p + q = 10 and that p and q are nonnegative. To find the maximum and minimum values of this expression, we can use the concept of optimization in algebra. First, express q in terms of p, using the given equation p + q = 10. We get q = 10 - p. Substitute this into the original expression to obtain a single-variable function f(p) = p2(10 - p)250. Now, we can differentiate this function with respect to p and find its critical points.

To find the difference between the maximum and minimum values, evaluate f(p) at the critical points and at the endpoints of the domain (since p and q have to be nonnegative, the domain is [0, 10]). The required difference is the greatest value of f(p) minus the smallest value of f(p), which is the solution to the student's problem. This can be found by tedious algebra and the final answer would be expressed as a fraction.

Answer 2

The difference between the maximum and minimum is

`\left(\displaystyle(5)/(63)\right)^2\left(\displaystyle(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(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(5)/(63)\right)^2\left(\displaystyle(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(5)/(63)\right)^2\left(\displaystyle(625)/(63)\right)^(250)`