question help (1 Viewer)

[horndog huang 2]

A cargo service operates by running a ship between port a and port b at a constant speed of v kilometres per hour

For a given v, the cost per hour of running the ship is 9000+10v^2 dollars.
Find the value of v which minimises the cost of the trip

ANS: 30

 

[BlueGas]

Brah I'm getting 9000, lol, what am I doing wrong. When you make the first derivative = 0, it'll be 20V = 0 therefore V = 0, and the second derivative is 20 which is greater than 0 so it's a minimum, to find C you sub v=0 into the first equation (9000+10v^2) and the answer is 9000. But that's what I'm getting haha.

 

[horndog huang 2]

haha thats what i did , but how can speed be 0km/hr

 

[BlueGas]

haha thats what i did , but how can speed be 0km/hr

The ship is not moving, haha, but for real I'm sure someone else will know how to work out this question.

 

[BlueGas]

How many parts are there to this question?

 

[horndog huang 2]

1 part

 

[horndog huang 2]

just to find minimum v

 

[psyc1011]

Since v is a real number, then v^2 is always positive and is at least 0.



That is the same as



Which means that C is at least 1000.

 

[Drsoccerball]

 

[Drsoccerball]

Since v is a real number, then v^2 is always positive and is at least 0.



That is the same as



Which means that C is at least 1000.

That depends on the distanc between A and B what if the distance is 0 ? it cost 0$...

 

[BlueGas]

How did you find v if you don't know the value of d?

 

[Drsoccerball]

How did you find v if you don't know the value of d?

We didnt have the value of d in this question....
In these types of questions the most important thing is
Edit: The D cancels in the derivative