A person coughs when a foreign object is in the windpipe. The velocity of the cough depends on the size of the object. Suppose a person has a windpipe with a 22-mm radius. If a foreign object has a radius r, in millimeters, then the velocity V, in millimeters/second, needed to remove the object by a cough is given by

V(r)= k(20r^2- r^3), 0< = r < =20

where k is some positive constant. For what size object is the maximum velocity needed to remove the object?

r = 13.33 m = (40/3) mm is the size of object for which the maximum velocity is needed to remove the object.

Step-by-step explanation:

v(r) = k(20r² - r³)

The function is at a maximum when (dv/dr) = 0 and (d²v/dr²) < 0

v(r) = (20kr² - kr³)

(dv/dr) = 40kr - 3kr²

At maximum v, (dv/dr) = 40kr - 3kr² = 0

40kr - 3kr² = 0

r (40k - 3kr) = 0

r = 0 or 40k - 3kr = 0

3kr = 40k

r = 40/3 = 13.333 mm

r = 0 mm or 13.33 mm

To investigate which point truly corresponds to the maximum value for the velocity,

(dv/dr) = 40kr - 3kr²

(d²v/dr²) = 40k - 6kr

At r = 0 mm

(d²v/dr²) = 40k - 6kr = 40k - 6k(0) = 40k

Since k is described as a positive constant, 40k > 0 and is positive,

Hence, (d²v/dr²) = 40k > 0,

r = 0 mm does not corresponds to a maximum point for the given function, rather, the conditions obtained show that it corresponds to a minimum point.

At r = 13.333 mm = (40/3) mm

(d²v/dr²) = 40k - 6kr = 40k - 6k(40/3) = 40k - 80k = - 40k.

Since k is described as a positive constant, - 40k < 0

Hence, (d²v/dr²) = - 40k < 0,

r = (40/3) mm corresponds to a maximum point for the given function, the conditions obtained show that it is indeed the value of r, when the velocity is at maximum value.

can you put the picture up?

step-by-step explanation: