differentiation problem????

Question

Find the least value of v+(6000/v) in the domain v>0

Answer 1

So..... What's the problem?? Take the 1'st and 2'nd derivatives of v + (6000/v). Set the 1'st derivative equal to 0 and solve for the points where dy/dv = 0. These are either maxima or minima. Put them into the 2'nd derivative and see if the 2'nd derivative at those points is negative or positive. If it's positive, the point is a minimum, if it's negative, the point is a maximum.

Doug

Answer 2

f(v) = v + 6000 v ^(-1)
f `(v) = 1 - 6000.v^(-2) = 0 for turning point
f " (v) = 12000.v^(-3)
6000.v^(-2) = 1
v² = 6000
v = √(400 x 15)
v = 20.√15
f "(20.√15) is +ve , therefore MINIMUM value of f(v) occurs at v = 20.√15
f (20.√15) = 20.√15 + 6000 / (20.√15)
f (20.√15) = [400 x 15 + 6000] / (20.√15)
f (20.√15) = 12000 / (20.√15)
f (20.√15) = 600 / √15
f (20.√15) = 600.√15 / 15
f (20.√15) = 40.√15 is least value.

Answer 3

Doug is right. But you dont really need to find the second derivative, just the sign of the first one will be enough

1 - 6000/v^2 = 0

1 = 6000/v^2

v^2 = 6000

v = +/-V6000

.............. ........ 0 ........... ..... 0
____+______|___-_|__-__|____+___>
................. - V6000... 0... V6000

The functions decreases until v is V6000, and then it grows. So, the least value is V6000

Ana