Find the absolute minimum

Question

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 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 below, where k is a constant and s is the normal radius of the trachea. The restriction of r is due to the fact that the trachea wall stiffens under pressure and a contraction greater than 0.5s is prevented (otherwise the person would suffocate).

v(r) = k(s - r)r^2
0.5s ≤ r ≤ s
(a) Determine the value of r in the interval [0.5s, s] at which v has an absolute maximum.

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

Answer 1

Hey Chris.

 

In this question, k and s are both constants, which will make the differentiation much easier. Let's distribute the terms to avoid having the use the product rule.

 

v(r) = (ks - kr)*r^2

v(r) = ks*r^2 - k*r^3

 

a) We want to find the absolute maximum, so we'll have to take the derivative and set it equal to 0.

 

v'(r) = 2ks*r - 3k*r^2

0 = 2ks*r - 3k*r^2

 

0 = r*(2ks - 3k*r)

 

r = 0, 2s/3

 

A radius of 0 falls outside of your interval (as it should be... if your trachea radius is 0 it means your throat has collapsed). So all we care about is r = 2s/3

 

Since we're looking for absolute maxima, we need to evaluate v(r) at the ends of the intervals as well as at 2s/r

 

v(0.5s) = k*(s - 0.5s)*(0.5s)^2

= k*(0.5s)*(0.25s^2)

 

= 0.125*k*s^3

 

v(2s/3) = k*(s - 2s/3)*(2s/3)^2

 

= k*(s/3)*(4s^2/9)

 

= 4/27*k*s^3

 

v(s) = k*(s - s)*s^2

 

= 0

 

The absolute maximum occurs at r = 2s/3

 

b)

 

Maximum value of v on the interval occurs where we just found the absolute max.

 

Max value:

 

v = 0.125k*s^3

 

With the information given, I don't believe it's possible to determine k. Though I may be missing something.

 

Hope this helps.