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 v of the airstream is related to the radius r of the trachea by the equation:
v(r) = k (r0 – r) r^2 , r0/2 greater than or equal to r greater than or equal to r0
where k is the constant and r0 is the normal radius of the trachea. The restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than r0 / 2 is prevented, otherwise the person would suffocate.
A) determine the value of r in the interval [r0 / 2 , r0 ] at which v has an absolute maximum
B) what is the absolute maximum value of v on the interval?
Please help and explain if work is not clear. use different numbers if you wish
