1. ## optimisation problem

Hello I have been stuck on this question and my maths exam is tomorrow. Ah.

(It gives no diagrams.)

a point M lies 500m north of point N.
Cyclist A starts from M, travelling south at 6m/s.
Cyclist B starts from N, moving east at 8m/s. Let D be the distance between the cyclists.
Let angle ABN = x.
a)show that D = 4000/(8sinx and 6 cosx)
b) calculate the shortest distance between the cyclists

I can do b) actually as that is simple differentiation and dy/dx = 0 etc. etc. but it is a), the part that equals the most marks which i am terribly stuck on.

2. Hello, mathshelpneeded!

This is not an easy one . . .

Point $\displaystyle M$ lies 500 m north of point $\displaystyle N$.
Cyclist $\displaystyle A$ starts from $\displaystyle M$, travelling south at 6 m/s.
Cyclist $\displaystyle B$ starts from $\displaystyle N$, moving east at 8 m/s.

Let $\displaystyle D$ be the distance between the cyclists.
Let $\displaystyle \angle ABN = x$

a) Show that: .$\displaystyle D \:= \:\frac{4000}{8\sin x + 6\cos x}$ . .
Why didn't they reduce this?

b) Calculate the shortest distance between the cyclists.
Code:
         M*
|
6t |
|
A*
| \
|   \
500-6t |     \D
|       \
|       x \
N* - - - - - *B
8t

Cyclist $\displaystyle A$ starts at $\displaystyle M$ and goes south at 6 m/s.
In $\displaystyle t$ seconds, he has gone $\displaystyle 6t$m to point $\displaystyle A$.
$\displaystyle AN \:= \:500-6t$

Cyclist $\displaystyle B$ starts at $\displaystyle N$ and goes east at 8 m/s.
In $\displaystyle t$ seconds, he has gone $\displaystyle 8t$m to point $\displaystyle B$.

We want the problem in terms of $\displaystyle x$, not $\displaystyle t$.

In right triangle $\displaystyle ANB\!:\;\;\tan x \:=\:\frac{500-6t}{8t}$

We have: .$\displaystyle 8t\tan x \:=\:500 - 6t\quad\Rightarrow\quad 8t\tan x + 6t \:=\:500$

Factor: .$\displaystyle 2t(\tan x + 3) \:=\:500\quad\Rightarrow\quad t \;=\:\frac{250}{4\tan x + 3}$ [1]

We also have: .$\displaystyle \cos x \:=\:\frac{8t}{D}\quad\Rightarrow\quad D \:=\:\frac{8t}{\cos x}$ [2]

Substitute [1] into [2]: .$\displaystyle D \:=\:\frac{8\left(\frac{250}{4\tan x + 3}\right)}{\cos x} \:=\:\boxed{\frac{2000}{4\sin x + 3\cos x}}$ (a)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .$\displaystyle D \:=\:2000(4\sin x + 3\cos x)^{-1}$

Then: .$\displaystyle D' \;= \;2000(-1)(4\sin x + 3\cos x)^{-2}(4\cos x - 3\sin x) \;=\;0$

We have: .$\displaystyle \frac{2000(3\sin x - 4\cos x)}{(4\sin x + 3\cos x)^2} \:=\:0\quad\Rightarrow\quad3\sin x-4\cos x \:=\:0$

Then: .$\displaystyle 3\sin x \:=\:4\cos x\quad\Rightarrow\quad\frac{\sin x}{\cos x} \:=\:\frac{4}{3}\quad\Rightarrow\quad \tan x \:=\:\frac{4}{3}$

Since $\displaystyle \tan x \;=\:\frac{4}{3} \;=\:\frac{opp}{adj}$, we have: .$\displaystyle opp = 4,\;adj = 3\quad\Rightarrow\quad hyp = 5$
. . Hence: .$\displaystyle \sin x = \frac{4}{5},\;\cos x = \frac{3}{5}$

Therefore: .$\displaystyle D \;=\;\frac{2000}{4\sin x + 3\cos x} \;=\;\frac{2000}{4(\frac{4}{5}) + 3(\frac{3}{5})} \;=\;\boxed{400\text{ m}}$ (b)

3. Originally Posted by mathshelpneeded
a)show that D = 4000/(8sinx and 6 cosx)
b) calculate the shortest distance between the cyclists
Here is a non-calculus solution to obtaining a maximum value of,
$\displaystyle D=\frac{2000}{4\sin x+3\cos x}$

First given,
$\displaystyle A\sin x+B\cos x$
Where $\displaystyle A,B$ are not both zero, we can write.
$\displaystyle \sqrt{A^2+B^2}\left( \frac{A}{\sqrt{A^2+B^2}}\sin x+\frac{B}{\sqrt{A^2+B^2}}\cos x \right)$.
Consider a right triangle.
With adjacent angle $\displaystyle \theta$ adjacent to leg $\displaystyle A$ and with opposite leg $\displaystyle B$. Then the hypotenuse is, $\displaystyle \sqrt{A^2+B^2}$.
And $\displaystyle \cos \theta = \frac{A}{\sqrt{A^2+B^2}}$ and $\displaystyle \sin \theta = \frac{B}{\sqrt{A^2+B^2}}$ with $\displaystyle \tan \theta = \frac{B}{A}$.
Thus,
$\displaystyle \sqrt{A^2+B^2}(\cos \theta \sin x+\cos x\sin \theta )=\sqrt{A^2+B^2}\sin (x+\theta )$
Thus,
$\displaystyle D=\frac{2000}{\sqrt{4^2+3^2}\sin (x+\theta )}=\frac{2000}{5\sin (x+\theta)}$
Note, the problem only asks to find the minimum value.
That happens when $\displaystyle \sin (x+\theta)$ is as large as possible thus $\displaystyle \sin (x+\theta)=1$.
Thus,
$\displaystyle D_{\mbox{max}}=\frac{2000}{5}=400$

4. Originally Posted by mathshelpneeded
Hello I have been stuck on this question and my maths exam is tomorrow. Ah.

(It gives no diagrams.)

a point M lies 500m north of point N.
Cyclist A starts from M, travelling south at 6m/s.
Cyclist B starts from N, moving east at 8m/s. Let D be the distance between the cyclists.
Let angle ABN = x.
a)show that D = 4000/(8sinx and 6 cosx)
b) calculate the shortest distance between the cyclists

I can do b) actually as that is simple differentiation and dy/dx = 0 etc. etc. but it is a), the part that equals the most marks which i am terribly stuck on.
I had this solution on my scratch paper last night before slept. Didn't have time to post it because it was almost 12:00 midnight.

Here it is, anyway.

I assumed D is line AB.

DsinX = 500 -6t -----------------(1)
DcosX = 8t -------------------(2)

From (1), t = (500 -DsinX)/6
From (2), t = (DcosX)/8
t = t,
(500 -DsinX)/6 = (DcosX)/8
Cross multiply,
8(500 -DsinX) = 6(DcosX)
4000 -8DsinX = 6DcosX
4000 = 6DcosX +8DsinX
4000 = D(8sinX +6cosX)
D = 4000 / (8sinX +6cosX) ------------shown.

----------------------
b) For the minimum D.

D = 4000 /(8sinX +6cosX)
Differentiate both sides with respect to X,
dD/dX = [(8sinX +6cosX)*0 -4000(8cosX -6sinX)] /[(8sinX +6sinX)^2]
Set dD/dX = 0,
0 = -4000(8cosX -6sinX)
0 = 8cosX -6sinX
6sinX = 8cosX
Divide both sides by cosX,
6tanX = 8
tanX = 8/6 = 53.13 degrees ----------when D is shortest.

Hence,
D = 4000 /(8sin(53.13deg) +6cos(53.13deg)) = 400m ------------answer.