Results 1 to 6 of 6

Math Help - Extrema exercise

  1. #1
    Junior Member
    Joined
    Dec 2009
    Posts
    51

    Extrema exercise

    Two high-ways cross at 60 angle. On both roads two cars are going at fixed speeds into the same direction, towards a crossing. One at 80 km/h, the other 100 km/h. At the beginning they were 10 km and 8 km away from the crossing. When the distance between the two cars will be the smallest?

    My attempt at solving:

    And it is not the correct answer, I'm supposed to get a little bit more than five minutes... Obviously the smallest distance between the cars will be when d^2 is the smallest..? Also, should I ignore the 30 angle or is it there for something I'll need?
    Also the function should be named d(t), not f(t).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78

    Re: Extrema exercise

    Quote Originally Posted by Evaldas View Post
    Two high-ways cross at 60 angle. On both roads two cars are going at fixed speeds into the same direction, towards a crossing. One at 80 km/h, the other 100 km/h. At the beginning they were 10 km and 8 km away from the crossing. When the distance between the two cars will be the smallest?

    My attempt at solving:

    And it is not the correct answer, I'm supposed to get a little bit more than five minutes... Obviously the smallest distance between the cars will be when d^2 is the smallest..? Also, should I ignore the 30 angle or is it there for something I'll need?
    Also the function should be named d(t), not f(t).
    First your error is assuming that the triangle is a right triangle. This is not true. This leads to the following question. Have you taken trigonometry? You would need to use the law of cosines to find the oblique triangle with the missing side as you have the cases (SAS).

    It would state that d^2=(10-80t)^2+(8-100t)^2-2(10-80t)(8-100t)\cos(60^\circ)

    using the fact that cosine of 60 degrees is one half gives

    d^2=(10-80t)^2+(8-100t)^2-(10-80t)(8-100t)

    This is a quadratic function in t and can be minimized by finding its vertex.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Dec 2009
    Posts
    51

    Re: Extrema exercise

    Isn't the shortest distance always the perpendicular?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123

    Re: Extrema exercise

    Quote Originally Posted by Evaldas View Post
    Isn't the shortest distance always the perpendicular?
    Perpendicular to what?

    (You've placed the right angle at the upper leg of the triangle. Why? Why not the lower leg?)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Dec 2009
    Posts
    51

    Re: Extrema exercise

    Quote Originally Posted by earboth View Post
    Perpendicular to what?

    (You've placed the right angle at the upper leg of the triangle. Why? Why not the lower leg?)
    In this case it would be the perpendicular from some point from the 80 km/h car trajectory into the 100 km/h (in other terms: I would draw a line touching the 100 km/h trajectory at an 90 degree angle coming from the 80 km/h path).
    And the angle is such because the 100 km/h car is going faster than the 80 km/h car and in the same amount of time the 100 km/h will go a bigger distance than the 80 km/h car... However, now that I think about it, one's path is 10 km and the other's is 8 km, so maybe that changes the angle...
    I'm confused.

    EDIT: Wait... The 100 km/h car would get to the crossing in 0.08h, and the 80 km/h - 0.125 h. So the perpendicular should be the opposite of what I did. Right?
    Last edited by Evaldas; November 11th 2011 at 06:33 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123

    Re: Extrema exercise

    Quote Originally Posted by Evaldas View Post
    In this case it would be the perpendicular from some point from the 80 km/h car trajectory into the 100 km/h (in other terms: I would draw a line touching the 100 km/h trajectory at an 90 degree angle coming from the 80 km/h path).
    And the angle is such because the 100 km/h car is going faster than the 80 km/h car and in the same amount of time the 100 km/h will go a bigger distance than the 80 km/h car... However, now that I think about it, one's path is 10 km and the other's is 8 km, so maybe that changes the angle...
    I'm confused.

    EDIT: Wait... The 100 km/h car would get to the crossing in 0.08h, and the 80 km/h - 0.125 h. So the perpendicular should be the opposite of what I did. Right?
    All your troubles are caused by your assumption that there must be a right angle somewhere.

    Please read TheEmptySet's reply thoroughly and you are only a few steps away from the final solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. An Exercise
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: January 15th 2011, 09:47 AM
  2. Exercise
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 25th 2009, 03:59 AM
  3. Absolute Extrema, Local Extrema
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 23rd 2009, 03:08 AM
  4. Replies: 1
    Last Post: November 18th 2009, 11:22 PM
  5. Help with an exercise
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 10th 2009, 11:35 PM

Search Tags


/mathhelpforum @mathhelpforum