- anonymous

A sherman is in a rowboat on a lake and 3
km from shore. He wishes to reach a store 2
km down the (straight) shore. He can row at
5 km/h and run at 13 km/h. To what point
down-shore should he row to get to the store
as quickly as possible?
It's supposed to be optimization but I'm wondering if I need to use related rates... :/

- jamiebookeater

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- anonymous

did you draw a picture?

- anonymous

i dont think you need to use related rates for this question

- anonymous

Yup gives me a right angle triangle |dw:1354488913281:dw|

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## More answers

- anonymous

the store being the point where 5 km/h and 13km/h meet

- anonymous

you drew it wrong

- anonymous

you cant use speed as distance values

- anonymous

im just showing the distance where the speed is going. I thought it was normal to do. ( well for me to understand)

- anonymous

|dw:1354489054440:dw|

- anonymous

|dw:1354489176738:dw|

- anonymous

Oh yeah ended the triangle too late. I need to find the the point where he reaches the shore,
But I usually get 2 equations, ( one is my constraint the other is the one I'm trying to maximize/minimize) I can't figure them out here
Perimeter, or area, I mean I feel as if I don't have enough variables.

- anonymous

d/v=t
|dw:1354489441544:dw|
first determine the rowing distance
\[d_{row}=\sqrt{x^2+9}\]
and running distance
\[d_{run}=2-x\]
now calculate total time to travel said distance
\[t= d_{row}/v_{row} + d_{run}/v_{run}\]
you should get a nice fun equation
then solve for x using first derivative
then solve for t

- anonymous

Ahhhhhh so its not always with area and peremeter.... stupid me. THanks a lot , I'll try it from there on my own :P Gotta love math

- anonymous

@MarcLeclair Have you found x yet?

- anonymous

No still trying hehe. Taking some time :(

- anonymous

Total time T = run time + row time
( 2- x ) / 13 + √ ( x² + 9 ) / 5
Can you take derivative:

- anonymous

yeah thats what im trying to do right now,
I got (-13x/13 ^2) +3(1/2( x ^2+9) ^(-1/2) (2x) / 9

- anonymous

er I dont mean 3 I mean 5 sorry got confuse in the numbers. so 5(1/2 * and the last one is /15

- anonymous

T' = x / 5√ ( x² + 9 ) - 1 / 13

- anonymous

Isn't mine right but not simplified?

- anonymous

T' = 0
x / 5√ ( x² + 9 ) = 1 / 13
5√ ( x² + 9 ) = 13x

- anonymous

Yes thank you chlorophyll :) I ll verify if my derivative was right. Thanks a lot for your help

- anonymous

It is pretty complicated, I've got more complicated to do haha. I'm enjoying it O_o

- anonymous

13x = 5√ ( x² + 9 )
-> 169x² = 25 ( x² + 9)
144x² = 225
x = 15/12

- anonymous

That's what I got too and it is right. You two were life saviour haha

- anonymous

=> x = 1hr 15'

- anonymous

its okay hahahaha, everybody have their strong points, you still helped with calculations :) I'm aweful at making relationships in math :P

- anonymous

Finally, plug x to find the distance, though :)
All credit should go to @completeidiot =)

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