Calculus Help

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?

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  1. I got this answer

    v = 4/27 kr^30

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  2. 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

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

    Thank you!

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  4. 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.

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