1. ## Optimizing time

I learned optimization (distance) but I missed the lesson on optimizing time so I'm having some trouble on this question:

A sailor in a boat 8 km offshore wants to reach a point on shore that is 10 km from the point that is directly opposite his current position in the shortest possible time. Find the landing point and traveling time if:
a) She rows at 4 km/h and runs at 6 km/h
b) She rows at 4 km/h and walks at 5 km/h (watch the Domain on this one)

I'm not sure how optimizing time differs from distance, so if someone could give me a quick explanation on part a I'll try part b. I know $\displaystyle time=\frac{distance}{velocity}$, but I'm not sure how to set up a relevant equation.

Thanks for the help!

2. Originally Posted by IanCarney
I learned optimization (distance) but I missed the lesson on optimizing time so I'm having some trouble on this question:

A sailor in a boat 8 km offshore wants to reach a point on shore that is 10 km from the point that is directly opposite his current position in the shortest possible time. Find the landing point and traveling time if:
a) She rows at 4 km/h and runs at 6 km/h
b) She rows at 4 km/h and walks at 5 km/h (watch the Domain on this one)

I'm not sure how optimizing time differs from distance, so if someone could give me a quick explanation on part a I'll try part b. I know $\displaystyle time=\frac{distance}{velocity}$, but I'm not sure how to set up a relevant equation.

Thanks for the help!
Always draw a diagram!

Using this we can we can find each distance and then use the formula
$\displaystyle t=\frac{d}{r}$

Use the Pythagorean theorem to find the length of the hypotenuse

Call this $d_1$ the other distance can be read of the diagram
$d_2=10-x$

So now the first time is

$\displaystyle t_1=\frac{d_1}{4}$ and

$\displaystyle t_2=\frac{d_2}{6}$

So the total time is

$T(x)=t_1+t_2$ now just minimize this problem!

3. Thanks a lot for the great explanation!

$\displaystyle T(x)=\sqrt{\frac{x^2+64}{4}} + \frac{10-x}{6}$

^Something like that, correct?

Also, do you know why it states to watch for the domain in part b?

4. Originally Posted by IanCarney
Thanks a lot for the great explanation!

$\displaystyle T(x)=\sqrt{\frac{x^2+64}{4}} + \frac{10-x}{6}$

^Something like that, correct?

Also, do you know why it states to watch for the domain in part b?
That is very close and maybe its just a typo but the 4 is not under the radical

$\displaystyle T(x)=\frac{\sqrt{x^2+64}}{4} + \frac{10-x}{6}$

As I haven't worked out the details of the 2nd part this is my guess

i) there is not any critical points in the interal $0 \le x \le 10$

If that is the cases where do the global max and mins occur?

5. Originally Posted by TheEmptySet
That is very close and maybe its just a typo but the 4 is not under the radical

$\displaystyle T(x)=\frac{\sqrt{x^2+64}}{4} + \frac{10-x}{6}$
Yes that's what I meant. I have another quick question if you don't mind, the answer to part a is 2.84 km, 3.15 hours. I know 3.15 is the $x$ value when the derivative is zero, but I'm not sure how they got 2.84 km. Are they asking for $d_1$? If so, how would I find that?

Thanks again!

6. Originally Posted by IanCarney
Yes that's what I meant. I have another quick question if you don't mind, the answer to part a is 2.84 km, 3.15 hours. I know 3.15 is the $x$ value when the derivative is zero, but I'm not sure how they got 2.84 km. Are they asking for $d_1$? If so, how would I find that?

Thanks again!
What is the relationship between "the landing point" and the x in your equation for T(x) ....?

7. Originally Posted by mr fantastic
What is the relationship between "the landing point" and the x in your equation for T(x) ....?
Landing point as in where she lands on the shore after rowing? And, I'm not sure:s

8. Originally Posted by IanCarney
Landing point as in where she lands on the shore after rowing? And, I'm not sure:s
Nearly all of the question has been done for you. You need to think some more about how the equations and the answers connect to your diagram and the questions.

9. Originally Posted by mr fantastic
Nearly all of the question has been done for you. You need to think some more about how the equations and the answers connect to your diagram and the questions.
I'm trying to find the distance of "x" in the diagram right?

10. Originally Posted by IanCarney
I'm trying to find the distance of "x" in the diagram right?
Do you understand what 'x' represents in 'your' solution ....? You need to take some ownership of the help (eg. the equation for T(x)) given to you so far.

11. Originally Posted by mr fantastic
Do you understand what 'x' represents in 'your' solution ....? You need to take some ownership of the help (eg. the equation for T(x)) given to you so far.
Ah, I get it now. I must have gotten confused over what T(x) meant. Thanks for your help!