A man launches his boat from point A on a bank of a straight river, 1 km wide, and wants to reach point B, 1 k?

A man launches his boat from point A on a bank of a straight river, 1 km wide, and wants to reach point B, 1 km downstream on the opposite bank, as quickly as possible. He could row his boat directly across the river to point C and then run to B, or he could row directly to B, or he could row to some point D between B and C and then run to B. If he can row 6 km/h and run 8 km/h, where should he land to reach B as soon as possible? (We assume that the speed of the water is negligible compared to the speed at which the man rows.) Find the answer in km for c.

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  • M3
    Lv 7
    1 decade ago
    Favorite Answer

    A

    |\

    |-\

    |--\

    |----\

    |-----\------

    C ...X ...B

    we shall solve in general form & then substitute numerical values

    let width of river AC = w,

    landing point X be x km from C

    & let the angle downstream (CAX) be z

    it can be shown that if running speed / rowing speed = k,

    then sin z = w/k will give the minimum time, which yields

    x = w/√(k^2-1)

    plugging in,

    x = 1/√((8/6)^2 - 1) = 3/√7 km > 1 km

    WHICH MEANS HE SHOULD ROW DIRECTLY TO B !!

    ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

    proof using #s given:

    ----------------------------

    dist. rowed = 1..sec z,

    dist. run, = 1 -1.tan z

    time t ∞ sec z - tan z / (4/3)

    dt/dz = sec z.tan z - 0.75sec^2 z

    = (sin z - 0.75) / cos^2 z

    for minima, sin z = 0.75

    the trignometric way yields a valuable insight,

    viz. the optimum angle downstream depends only on k

    ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

    however, if you prefer the other method,

    x/6√(x^2+1) =1/8

    √(x^2+1)=(4/3) x

    x^2+1 = 16x^2/9

    7x^2 = 9

    x = 3/√7

  • cidyah
    Lv 7
    1 decade ago

    A

    |\

    |-\

    |--\

    |----\

    C---\------B

    ------x------

    CB=1

    AC=1

    Distance rowed = sqrt(x^2+1)

    Speed of rowing = 6

    Time = sqrt(x^2+1)/6

    Distance run = 1-x

    Speed of running = 8

    Time running = (1-x)/8

    Let s = sqrt(x^2+1)/6 + (1-x) / 8 --- total time to reach from A to B

    We want to minimize s with respect to x

    ds/dx = 2x/12sqrt(x^2+1) -1/8 = 0

    x/6sqrt(x^2+1) =1/8

    6 sqrt(x^2+1) = 8x

    sqrt(x^2+1)=(8/6) x

    x^2+1 = 4x/3

    3x^2+3=4

    3x^2=1

    x^2=1/3

    x=1/sqrt(3) =0.578 km

  • 4 years ago

    you desire to optimize the cost. So variety a equation for what the cost of the cable must be. cost = 0.10 x length of cable to abode + 0.07 x length of cable alongside abode. you may desire to persist with some geometry so as to be sure what the dimensions of cable to abode and length of cable alongside abode may be. there's a dating.

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