# It's in my Exam|Optimization Question|Picture Included

1. Mar 22, 2007

### Raza

1. The problem statement, all variables and given/known data
Ship B is going west at 12km\h[W] while Ship A is going south at 9km\h. In the beginning, the distance that they are away from each other is 75km. When will the ships be the closest?

2. The attempt at a solution
$$D^2=(75-12t)^2+(9t)^2$$

$$D^2=5625-1800t-144t^2+81t^2$$

$$D^2=5625-1800t-63t^2$$

$$\frac{dD}{dt}=-1800-126t$$

$$0=-1800-126t$$

$$t=\frac{1800}{-126}$$

And t can't be a negative number.

Last edited: Mar 23, 2007
2. Mar 22, 2007

### e(ho0n3

Your diagram is wrong. You have ship A going south and ship B going west. Anywho, to solve this problem, determine an equation that will give you the distance D between ship A and ship B as observed from one of the ships.

3. Mar 22, 2007

### Raza

$$D=\sqrt{(75)^2+(x)^2}$$

$$D=\sqrt{5625+x^2}$$

then what?

Last edited: Mar 23, 2007
4. Mar 22, 2007

### e(ho0n3

How are you deriving those equations?

5. Mar 22, 2007

### e(ho0n3

This equation was better except for an error in the first term on the RHS.

6. Mar 22, 2007

### Raza

I saw those equation used in a textbook in a similar question.

What's RHS?

And what do I now after setting up the equation for D?
$$D=\sqrt{(75)^2+(x)^2}$$

$$D=\sqrt{5625+x^2}$$

$$\frac{dD}{dt}=\frac{x}{\sqrt{5625+x^2}}?$$

Is this what I do?

Last edited: Mar 23, 2007
7. Mar 23, 2007

### e(ho0n3

It means right hand side.

You have not explained how you derived that equation.

8. Mar 23, 2007

### Raza

It's the pythagorean theorem.

9. Mar 23, 2007

### e(ho0n3

OK then. Next question: Why did you decide do use it? How are you using it? Does it make sense?

10. Mar 23, 2007

### Raza

Okay, I got the answer.

$$D^2=(75-12t)^2+(9t)^2$$

$$D^2=5625-1800t+144t^2+81t^2$$ *Note that it's +144t2, not -144t2 like before. Just a multiplication error.:uhh:

$$D^2=5625-1800t+225t^2$$

$$\frac{dD^2}{dt}=-1800+450t$$

$$0=-1800+450t$$

$$t=\frac{1800}{450}$$

$$t=4$$

Therefore, The ship will be the closest at 4 hours with the distance of 45km apart.

Last edited: Mar 23, 2007