# Math Help - optimzation problem #8

1. ## optimzation problem #8

A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?

totally stuck on this one.

2. Originally Posted by skeske1234
A train leaves the station at 10:00 and travels due south at a speed of 60km/h. Another train has been heading due west at 45 km/h and reaches the same station at 11:00. At what time were the two closest together?

totally stuck on this one.
let $d_1$ = distance first train is from the origin (station)

$d_1 = 60t$

$d_2$ = distance second train is from the same station

$d_2 = 45 - 45t$

let $r$ = straight line distance between the two trains between 10:00 and 11:00

$r^2 = d_1^2 + d_2^2$

minimize $r^2$ and $r$ will also be minimized.

3. Originally Posted by skeeter
let $d_1$ = distance first train is from the origin (station)

$d_1 = 60t$

$d_2$ = distance second train is from the same station

$d_2 = 45 - 45t$

let $r$ = straight line distance between the two trains between 10:00 and 11:00

$r^2 = d_1^2 + d_2^2$

minimize $r^2$ and $r$ will also be minimized.
Ok, can you or someone check my work below because I can't seem to get the answer with the back.

I said..

Let x be the distance between A and B in km
Let a and b be the distance travelled by A and B in km

Min x at t=?
V=d/t
For A
a=60t
For B
b=45t

b^2+a^2=x^2
(45t)^2+(60t)^2=x^2
x=(5625t^2)^0.5
x'=5625t/[(5625t^2)^0.5]

x'=0 when t=0

domain of t
10<=t<=11

x(10)=750 km
x(11)=825 km

therefore the min distance between train A and train B is 750 km at 10:00

BUT>>> the back of the book's answer is: t=0.36 h

How do i get this and what did I do wrong?

4. Originally Posted by skeske1234
Ok, can you or someone check my work below because I can't seem to get the answer with the back.

I said..

Let x be the distance between A and B in km
Let a and b be the distance travelled by A and B in km

Min x at t=?
V=d/t
For A
a=60t
For B
b=45t

b^2+a^2=x^2
(45t)^2+(60t)^2=x^2 this equation is wrong
take another look at my original post ...

$r^2 = (60t)^2 + (45 - 45t)^2
$

minimizing $r^2$ will minimize $r$ ...

take the derivative of the RHS and set = 0

$2(60t)(60) + 2(45-45t)(-45) = 0$

$3600t - 2025 + 2025t = 0$

$5625t = 2025$

$t = \frac{2025}{5625} = \frac{9}{25} = .36$ hr

5. im still not sure with the distance of train 2 formula

you have 45-45t =d_2
I don't really understand this... why do we take 45 away from 45t? or the speed away from the distance?

6. Originally Posted by skeske1234
im still not sure with the distance of train 2 formula

you have 45-45t =d_2
I don't really understand this... why do we take 45 away from 45t? or the speed away from the distance?
the distances, $d_1$ and $d_2$ are measured from the origin.

the first train starts at the origin and moves away from the origin at 60 km/hr

$d_1 = 60t$

the second train started 45 km away from the origin, and heads back toward the origin at 45 km/hr.

$d_2 = 45 - 45t$

you see, after 1 hr, $d_2 = 45 - 45(1) = 0$ km from the origin. it made it back.