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Math Help - Application: Maxima and Minima

  1. #1
    Junior Member ihmth's Avatar
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    Application: Maxima and Minima

    1.) A man in a motorboat at A, receives a text message at noon, calling him to B. A bus making 40 mph leaves C, bound for B, at 1:00 pm. If AC = 30 mi, what must be the speed of the boat, to enable the man to catch the bus?

    2.) The lower edge of a picture is "a ft", the upper edge is "b ft", above the eye of an observer. At what horizontal distance shall he stand, if the vertical angle subtended by the picture is the greatest.

    Can anyone give me the right solution for these problems? Especially for number 1--I dont know where A, B, and C are located and I can't find the right solution .

    -tnx
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  2. #2
    Super Member

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    Hello, ihmth!

    I don't understand the diagram for the first one . . .


    2) The lower edge of a picture is a ft, the upper edge is b ft above the eye of an observer.
    At what horizontal distance shall he stand
    if the vertical angle subtended by the picture is the greatest?
    I have a rather unplesant solution . . .
    Code:
                                  * P
                                * | : 
                              *   | :
                            *     | :
                          *       | b-a
                        *         | : 
                      *           | :
                    *             | :
                  *               * Q
                *           *     | :
              * θ     *           | a
            *   *                 | :
        E * - - - - - - - - - - - * R
                      x

    E is the eye of the observer.
    PQ is the height of the picture.
    . . a = QR,\;\;b = PR,\;\;PQ \:=\: b-a

    Let x = ER, the horizontal distance.
    Let \theta = \angle PEQ

    In right triangle QRE\!:\;\;EQ \:=\:\sqrt{x^2+a^2}

    In right triangle PRE\!:\;\;EP \:=\:\sqrt{x^2+b^2}


    In \Delta PEQ, use the Law of Cosines: . \cos\theta \:=\:\frac{EQ^2 + EP^2 - PQ^2}{2(EQ)(EP)}

    We have: . \cos\theta \;=\;\frac{(x^2+a^2) + (x^2+b^2) - (b-a)^2}{2\sqrt{x^2+a^2}\sqrt{x^2+b^2}}

    . . which simplifies to: . \cos\theta \;=\;\frac{x^2+ab}{\sqrt{(x^2+a^2)(x^2+b^2)}}


    And I let you maximize \theta . . .


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

    I have an alternate equation: . \tan\theta \;=\;\frac{(b-a)x}{x^2+ab}

    . . but its derivation is even harder to explain.

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