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Math Help - Applications of Maxima/Minima

  1. #1
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    Applications of Maxima/Minima

    An island is at point A, 4km offshore from the nearest point B on a straight beach. A woman on the island wishes to go to a point C, 6km down the beach from B. She can go by rowboat at 5km/hr to a point P Between B and C. And then walk at 8km/hr along a straight path from P to C.
    a.) Estimate on your graphics calculator the route from A to C that she can travel in the least time.
    b.) Confirm Your estimates in part a analytically.
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    An island is at point A, 4km offshore from the nearest point B on a straight beach. A woman on the island wishes to go to a point C, 6km down the beach from B. She can go by rowboat at 5km/hr to a point P Between B and C. And then walk at 8km/hr along a straight path from P to C.
    a.) Estimate on your graphics calculator the route from A to C that she can travel in the least time.
    b.) Confirm Your estimates in part a analytically.
    Me, analytically only.

    The question asks for route for the least time, so we find the distance for the least time.

    distance = rate*time,
    So, time = distance / rate -----------**

    Rates are given, so we find distances.

    Let x = BP, in km.
    So, 6-x = PC, in km.

    In right triangle ABP, by Pythagorean theorem,
    AB = sqrt(4^2 +x^2) = sqrt(16 +x^2)

    Then, the times,
    Let T = total time spent from A to P to C,
    T = [sqrt(16 +x^2) / 5] +(6-x)/8 -------------------------(i)
    Differentiate both sides with respect to x,
    dT/dx = (1/5)[(1/2)(2x /sqrt(16 +x^2))] +(1/8)[-1]
    dT/dx = (1/5)[x /sqrt(16 +x^2)] -1/8
    Equate dT/dx to zero,
    0 = (1/5)[x/sqrt(16 +x^2)] -1/8
    Clear the fractions, multiply both sides by 5*8*sqrt(16 +x^2).
    0 = 8x -5sqrt(16 +x^2)
    sqrt(16 +x^2) = 8x/5
    sqrt(16 +x^2) = 1.6x
    Square both sides,
    16 +x^2 = (2.56)x^2
    16 = (1.56)x^2
    x^2 = 16/1.56 = 10.2564
    x = sqrt(10.2564) = 3.20256 km -----------for least T.

    Hence, for the route, or distance of the route, when x = 3.20256 km,
    D = sqrt(16 +x^2) +(6-x)
    D = sqrt(16 +10.2564) +(6 -3.20256)
    D = 7.92 km. ---------------------------answer.
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    An island is at point A, 4km offshore from the nearest point B on a straight beach. A woman on the island wishes to go to a point C, 6km down the beach from B. She can go by rowboat at 5km/hr to a point P Between B and C. And then walk at 8km/hr along a straight path from P to C.
    a.) Estimate on your graphics calculator the route from A to C that she can travel in the least time.
    ...
    Hello,

    to a)
    To estimate the time needed take the most "simplest" ways she can go:

    i) directly from A to C. The distance AC = √(16+36) ≈ 7.211 km. For this distance she needs 1 h; 26 min; 32 s

    ii) from A to B and from B to C. The time needed is 4/5 h + 6/8 h = 31/20 h = 1 h; 33 min;

    That means the minimum of time must be smaller than 1 h; 26 min; 32 s

    With ticbol 's result she needs 1 h; 22 min; 28 s

    EB
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