1. ## 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).

2. ## Re: Extrema exercise

Originally Posted by Evaldas
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.

3. ## Re: Extrema exercise

Isn't the shortest distance always the perpendicular?

4. ## Re: Extrema exercise

Originally Posted by Evaldas
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?)

5. ## Re: Extrema exercise

Originally Posted by earboth
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?

6. ## Re: Extrema exercise

Originally Posted by Evaldas
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.