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.

y=mx+

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .