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October 1, 2014

October 1, 2014

Posted by **Selda** on Friday, April 4, 2014 at 10:52pm.

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?

- Calculus Help -
**Selda**, Friday, April 4, 2014 at 10:56pmI got this answer

v = 4/27 kr^30

- Calculus Help -
**Steve**, Friday, April 4, 2014 at 11:00pmv = 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

- Calculus Help -
**Selda**, Friday, April 4, 2014 at 11:04pmI plugged in that answer but it marked it wrong. Are you sure that's answer for this question?

Thank you!

- Calculus Help -
**Steve**, Friday, April 4, 2014 at 11:06pmThat'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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