Question

It costs 32000+2v^3 dollars to operate a cruise ship for an hour, where v is the velocity
of the ship measured in kilometres per hour. We want to operate the ship from Victoria
to Seattle, which are 120 kilometres apart. How long will it take us to do this trip if
we operate choosing the velocity that minimizes the total cost?

Warning: When you find the point you are looking for, make sure you actually explain
why it is a maximum/minimum of the function you are studying. It costs 32000+2v^3 dollars to operate a cruise ship for an hour, where v is the velocity
of the ship measured in kilometres per hour. We want to operate the ship from Victoria
to Seattle, which are 120 kilometres apart. How long will it take us to do this trip if
we operate choosing the velocity that minimizes the total cost?

Warning: When you find the point you are looking for, make sure you actually explain
why it is a maximum/minimum of the function you are studying. @Mathematics

Answer 1

it will take infinity to operate this trip if you operate choosing the velocity that minimizes the total cost...

Answer 2

because the minimum velocity tends to zero.............

Answer 3

Find the derivative first.

Answer 4

finding derivative and equating it with zero gives velocity zero...

Answer 5

@saruz
Obviously v>0 is most cost effective.
since you cant generate any revenue with a cruise ship that just sits in port.

Answer 6

the derivative of 32000+2v^3?

Answer 7

v tends to zero , so time tends to infinity..........

Answer 8

i dont mean v equal to zero

Answer 9

Yeah i think so, it looks similar to a problem that i did.

Answer 10

link to your problem @ Mimi_x3

Answer 11

lols, its not online xD
I will have to find it first wait

Answer 12

@saruz
obviously 0=6v^2
v=0
but, we know that cant be true, and we know that v has to be postive: so you take an infitesimal value for v.

Answer 13

i dont mean v = 0, v is very near to zero but not zero

Answer 14

um, is there a typo with "32000+2v^3" I think that it should be "32000v + 2v^3" it might make more sense

Answer 15

ya. that makes sense

Answer 16

now it is worth solving with derivative

Answer 17

oh..but the question is 32000+2v^3

Answer 18

Yeah, but it won't make any sense with 3200, if you take the derivative of that its 0 xD

Answer 19

take 0.0000000000001
as your v.
evaluate.
Reasoning, v must be more than 0;
So, you substituted for an infitesimal and you tipped you're hat to Newton and Leibniz while you were hanging out in the 1600s

Answer 20

change you're to your
unless you really are a hat.

Answer 21

dont need any calculation, just reasoning is fine

Answer 22

lol i am doing this same homework.

I didnt really know how to approach it so i just said the cost is (32000+2v^3)*d/v since the time = distance/velocity. then i just took the derivative and found the minimum

Personally i am stuck on the question after this about finding absolute errors or some such.

Answer 23

are u guyz from same school...

Answer 24

oh cool ><

Answer 25

I think that its 32000v xD

Answer 26

I think what I said is right.
Saruz is technically correct about minimization, but his Lemma fails over the word problem because it requires movement to evaluate.
Mimi_x3 has a problem that makes sense, but isn't the problem.

Answer 27

Well, the problem will make sense and solvable with 3200v, i have a problemm that is similar to that