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Math Help - optimisation problem

  1. #1
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    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.
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  2. #2
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    Hello, mathshelpneeded!

    This is not an easy one . . .


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

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

    a) Show that: . 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 A starts at M and goes south at 6 m/s.
    In t seconds, he has gone 6tm to point A.
    AN \:= \:500-6t

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

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


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

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

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


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

    Substitute [1] into [2]: . 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: . D \:=\:2000(4\sin x + 3\cos x)^{-1}

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

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


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

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  3. #3
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    Quote Originally Posted by mathshelpneeded View Post
    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,
    D=\frac{2000}{4\sin x+3\cos x}

    First given,
    A\sin x+B\cos x
    Where A,B are not both zero, we can write.
    \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 \theta adjacent to leg A and with opposite leg B. Then the hypotenuse is, \sqrt{A^2+B^2}.
    And \cos \theta = \frac{A}{\sqrt{A^2+B^2}} and \sin \theta = \frac{B}{\sqrt{A^2+B^2}} with \tan \theta = \frac{B}{A}.
    Thus,
    \sqrt{A^2+B^2}(\cos \theta \sin x+\cos x\sin \theta )=\sqrt{A^2+B^2}\sin (x+\theta )
    Thus,
    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 \sin (x+\theta) is as large as possible thus \sin (x+\theta)=1.
    Thus,
    D_{\mbox{max}}=\frac{2000}{5}=400
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  4. #4
    MHF Contributor
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    Quote Originally Posted by mathshelpneeded View Post
    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.
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