# When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward causing an increase in pressure in the lungs. This is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to flow through. For a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. The greater the velocity of the airstream, the greater the force on the foreign object. X rays show that the radius of the circular tracheal tube contracts to about two-thirds of its normal radius during a cough. According to a mathematical model of coughing, the velocity v of the airstream is related to the radius r of the trachea by the equation

v(r) = k(r0 − r)r^2(1/2)r0 ≤ r ≤ r0

where k is a constant and r0 is the normal radius of the trachea. The restriction on r is due to the fact that the trachea wall stiffens under pressure and a contraction greater than

1/2r0 is prevented (otherwise the person would suffocate).

(b) What is the absolute maximum value of v on the interval?

## I got this answer

v = 4/27 kr^30

## v = k(r0-r)r^2 = kr0*r^2 - kr^3

dv/dr = 2kr0*r - 3kr^2 = kr(2r0 - 3r)

so, v obtains a max at r = 2/3 r0

what about the ends of the interval?

v(1/2 r0) = k(1/2 r0)(1/4 r0^2) = k/8 r0^3

v(2/3 r0) = k(1/3 r0)(4/9 r0^2) = 4k/27 r0^3

v(r0) = 0

so, since 4/27 > 1/8 max v = 4k/27 r0^3

## That's what I get. It seems we agree. Is there a chance the answer key is in error?

Maybe Reiny or Damon can see an error.

## I plugged in that answer but it marked it wrong. Are you sure that's answer for this question?

Thank you!

## To find the absolute maximum value of v on the interval, we need to find the value of r that maximizes the function v(r) = k(r0 - r)r^2(1/2), where r0 is the normal radius of the trachea.

To do this, we can take the derivative of v with respect to r and set it equal to zero to find the critical points:

dv/dr = k(r0 - r)(2r)(1/2) + k(r0 - r)r^2(1/2)(-1/2) = 0

Simplifying the expression:

k(r0 - r)(2r)(1/2) - k(r0 - r)r^2(1/2)/2 = 0

Multiplying through by 2*(2r)^(-1/2) to eliminate the fraction:

k(r0 - r)r^(1/2) - k(r0 - r)r = 0

Expanding and rearranging terms:

2(r0 - r)r^(1/2) - 2(r0 - r)r = 0

Distributing the 2:

2r0r^(1/2) - 2rr^(1/2) - 2r0r + 2r^2 = 0

Combining like terms:

2r0r^(1/2) - 2rr^(1/2) - 2r0r + 2r^2 = 0

Factor out r^(1/2):

r^(1/2)(2r0 - 2r) - (2r0 - 2r) = 0

Factor out (2r0 - 2r):

(2r0 - 2r)(r^(1/2) - 1) = 0

Setting each factor equal to zero:

2r0 - 2r = 0 or r^(1/2) - 1 = 0

Solving the first equation:

2r0 = 2r

r0 = r

This means that when r0 = r, the derivative becomes zero. However, this does not satisfy the inequality restriction r <= r0, so we can discard this critical point.

Solving the second equation:

r^(1/2) - 1 = 0

Taking the square of both sides:

r - 1 = 0

r = 1

Since the only critical point that satisfies the inequality restriction is r = 1, this is the only critical point we need to consider.

Now, we need to evaluate v at the critical point and at the endpoints of the interval r0 <= r <= r0:

v(1) = k(r0 - 1)(1^2)^(1/2) = k(r0 - 1)(1) = k(r0 - 1)

v(r0) = k(r0 - r0)r0^2(1/2) = 0

Since we can see that v(1) > v(r0) for any value of k and r0, the absolute maximum value of v on the interval is v(1) = k(r0 - 1).