# When a foreign object lodged in the trachea (windpipe) forces a person to cough, the diaphragm...

## 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 then 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 {eq}v {/eq} of the airstream is related to the radius {eq}r {/eq} of the trachea by the equation: {eq}v(r) = k (r_0 - r) r^2 , \frac{r_0}2 \geq r \geq r_0 {/eq} where {eq}k {/eq} is the constant and {eq}r_0 {/eq} is the normal radius of the trachea. The restriction on {eq}r {/eq} is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than {eq}\frac{r_0}2 {/eq} is prevented, otherwise the person would suffocate.

A) Determine the value of r in the interval {eq}[\frac{r_0}2 , r_0 ] {/eq} at which {eq}v {/eq} has an absolute maximum.

B) What is the absolute maximum value of {eq}v {/eq} on the interval?

## Absolute Maximum:

The maximum values of a function over an interval are determined:

from the critical points of the interval

and the ends of the interval

Among them is the value where the absolute extremum of the function is reached.

To determine the absolute maximum of the function over the interval, first, we need to determine the critical points of the interval:

{eq}v(r) = k({r_0} - r){r^2},\quad \frac{{{r_0}}}{2} \le r \le {r_0}\\ \\ v'(r) = k{r^2} + k({r_0} - r)2r\\ v'(r) = kr\left( {r + 2({r_0} - r)} \right)\\ v'(r) = kr\left( {r + 2{r_0} - 2r} \right)\\ v'(r) = kr\left( {2{r_0} - r} \right) = 0\\ \\ \left\{ \begin{array}{l} r = 0\\ 2{r_0} - r = 0 \to r = 2{r_0} \end{array} \right. {/eq}

According to the results we don't have critical points over the interval. So, the absolute maximum will be reached at the ends of the interval:

{eq}\left\{ \begin{array}{l} v(\frac{{{r_0}}}{2}) = k({r_0} - \frac{{{r_0}}}{2}){\left( {\frac{{{r_0}}}{2}} \right)^2} = k{\left( {\frac{{{r_0}}}{2}} \right)^3}\\ v({r_0}) = k({r_0} - {r_0}){r^2} = 0 \end{array} \right. {/eq}

A) So, the value of {eq}r {/eq} in the interval at which the {eq}v {/eq} has absolute maximum is {eq}r = \frac{{{r_0}}}{2}. {/eq}

B) For this value of the variable {eq}r {/eq}, the maximum value of the function is {eq}v\left( {\frac{{{r_0}}}{2}} \right) = k{\left( {\frac{{{r_0}}}{2}} \right)^3} {/eq}