**0**

# Differentiation Problem!

### #1

Posted 08 October 2006 - 11:45 AM

4. A ship has a 200 Km journey to make at a constant speed. At x km/h the cost of the journey will be

(x squared + 4000/x) per hour.

a. Find an expression for the time taken for the journey

the total cost of the trip.

Im completely lost with part a.

b. Find the speed that minimises the cost of the journey and calculate this minimum cost.

I got as far to find x = 12.6. Is this right?

However when I do the table I get a max value rather than a minimum value.

Ive got the same problem with Q5 a and b. Any ideas?

### #3

Posted 08 October 2006 - 05:07 PM

^{2}+ 4000/x) - Its all inside one bracket. Just the 4000 is divided by x.

### #4

Posted 08 October 2006 - 05:30 PM

For (b), you want to differentiate the cost and set equal to zero. Solving for

*x*gives you the stationary values. When differentiating , you need to write the second term with a negative power first.

You might have differentiated wrongly. I got .

Hope this helps.

### #5

Posted 08 October 2006 - 05:39 PM

Maybe not!

### #6

Posted 08 October 2006 - 05:49 PM

I see what you've done, you've multiplied everything by

*x*, is that right?

The reason you can't do that is because

is not an correct equation, the two things are not equal (think about their graphs). You can only put an equals symbol between two things that have the same value!

This is like saying that if the temperature of water after being heated is given by then then temperature is given by "1", which it's clearly not!

You see?

### #7

Posted 08 October 2006 - 05:54 PM

Ive got as far as x = 3rd root 2000 like you had. But when I sub this into (2x - (4000/ x squared)) it doesnt seem to work.

### #8

Posted 08 October 2006 - 05:55 PM

### #9

Posted 08 October 2006 - 05:57 PM

Yea 5 is just as hard.

### #10

Posted 08 October 2006 - 06:57 PM

*C*.

When , say , .

When , say , .

This gives:

For Q5, you know that the volume cm

^{3}. But you also know a way to work this volume out.

The volume of the cylinder is given by , and whatever this is must be equal to 144.

Does that help?

### #11

Posted 08 October 2006 - 07:00 PM

Yea that helps Steve - you are a maths genius!

What do they sub into what?

### #12

Posted 08 October 2006 - 07:20 PM

### #13

Posted 08 October 2006 - 07:22 PM

The time taken is just given by , both of which you are given (the distance is 200 and the speed is

*x*).

Then the cost is just the time multiplied by the cost per hour. I get that this is . Differentiating and solving equal to zero gives you the answer in the book. Then all you have to do is put this speed into your formula for cost.

I'm really sorry, I stupidly took the given equation as the total cost. Oh well, at least you've had lots of practice at differentiating!

### #14

Posted 08 October 2006 - 07:32 PM

So 4a. T = D/S = 200/x

How do you get the cost per hour for the second part?

### #15

Posted 08 October 2006 - 07:33 PM

### #16

Posted 08 October 2006 - 07:42 PM

Im still not sure!

So what is the working for part b? I cant do part 2 of a so how would you begin with the second part?

### #17

Posted 08 October 2006 - 08:27 PM

You should end up with what I said above.

### #18

Posted 09 October 2006 - 09:04 PM

#### 1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users