# Thread: Optimisation Problem?

1. ## Optimisation Problem?

Vehicle A is 25km due east of vehicle B. A moves due west at 60km/h and B moves due north at 80km/h. Find an expression for the distance of separation for the two vehicles t hours later. After how many minutes is this separation distance a minimum and what is this minimum distance?

Im really just not sure how to find the expression, but i think i know how to find at how many minutes the separation distance is minimum (differentiate the expression then make f'(x) = 0. Right?)

I found the expression for distance as d = 25 + 85x, but this is wrong. Apparently (according to my teacher) i need to use the Pythagoras theorem, so i guess that means im finding displacement and not distance.... now that i reread the question thats actually pretty obvious because its distance of separation not distance travelled. :/. oh well.

Anywho, any help in finding that expression is appreciated. Thanks.

2. Distance equals rate times time, right?. So, after t hours A have moved

$25-60t$ km and B has moved $80t$ km.

The distance between them is, by ol' Pythagoras,

$D^{2}=(25-60t)^{2}+(80t)^{2}$

Now, can you finish?.

3. Im unsure how to differentiate this (very new to it- just started intro-calc a couple of weeks ago) but i think im on the right track.

s = √((25-60t)² + (80t)²)
s = √(625 - 3000t + 10000t²)
s = 25 - 54.77t^(1/2) + 100t
s' = -27.58/(√t) + 100
0 = -27.58/(√t) + 100
100√t = 27.58
t = 0.075 h
t = 4.5 min

but this is wrong, the answer is double this - its 9. Im really not sure what im not getting, this question has been bugging me all night. :/

thankyou for your help

4. We can differentiate with out the radical to make it easier.

Expand $(25-60t)^{2}+(80t)^{2}=1000t^{2}-3000t+625$

Differentiate: $20000t-3000=20t-3$

$20t-3=0$

$t=\frac{3}{20}=.15$

$60(\frac{3}{20})=9$

Plug into Pythagoras to find the distance.