# Applied min and max!

• Apr 7th 2010, 05:12 PM
hotdogking
Applied min and max!
A girls kite has come down into a nearby field. The girl standing on a straight road wants to run down the road and then run diagonally through the field to her kite. The kite came down 1 mile to the closest point on the road, and she is presently 2 miles from that closest point. She can run 5 mph on the road and 3 mph in the field. At what point should she leave the road and cut across the field on order to get to her kite as quickly as possible??

really confused at how to setup and solve this one? really need help!
• Apr 7th 2010, 07:56 PM
eddie2042
Quote:

Originally Posted by hotdogking
A girls kite has come down into a nearby field. The girl standing on a straight road wants to run down the road and then run diagonally through the field to her kite. The kite came down 1 mile to the closest point on the road, and she is presently 2 miles from that closest point. She can run 5 mph on the road and 3 mph in the field. At what point should she leave the road and cut across the field on order to get to her kite as quickly as possible??

really confused at how to setup and solve this one? really need help!

See the attached picture and tell me how it goes. Do you have the answers with you? Please reply =D

Attachment 16259
• Apr 7th 2010, 08:00 PM
Nicholas
Quote:

Originally Posted by eddie2042
See the attached picture and tell me how it goes. Do you have the answers with you? Please reply =D

Attachment 16259

+1 for this method.
• Apr 8th 2010, 04:49 AM
hotdogking
yeah after seting it up the same way, i got the same! thanks
• Apr 7th 2013, 09:57 AM
Jhnny025
Re: Applied min and max!
I have a question on your derivative. If T = 5a + 3*sqrt(5-4a+a^2), you said the derivative would be 5 + 3*((2a-4)/sqrt(5-4a+a^2)). What happened to the exponential chain? Since T has a sqrt in the first place, the exponent would be (5-4+a^2)^(1/2). For the derivative, wouldn't you have to bring down the (1/2) exponent? Maybe I'm just missing something. Please let me know. Thx!
• Apr 14th 2013, 05:07 AM
eddie2042
Re: Applied min and max!
Wow looking back at this, I did the whole thing wrong. The time equation is supposed to be

$\displaystyle T = \frac{a}{5} + \frac{b}{3}$

By redoing the whole thing (and yes you're right about bringing the 1/2 down), I got a = b = 1.25 miles.

...sorry.