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Math Help - The Minimum Value of The Sum

  1. #1
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    Post The Minimum Value of The Sum

    Hello everyone I have a question asking the minimum value for the sum of (|AP|+|PD|)How can I solve this can you explain step by step? Thanks for your contribution. Please have a look at the attachment.

    PS:I have found the solution by writing two seperate cosinus theorem functions adding them together than taking the derivative of the resulting function, equaling the derivative function to "0" and substituting the root with x in the main function that gives the sum of |AP|+|PD| with the help of TI-89 Titanium but this exam must be solved without the aid of any kind of calculator. The Correct answer is \sqrt241
    Attached Thumbnails Attached Thumbnails The Minimum Value of The Sum-tarama0001.jpg  
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  2. #2
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    Hello, JohnDoe!

    Your game plan is excellent.
    It can be solved without a calculator,
    . . but it takes a lot of Algebra . . .


    Locate P so that |AP| + |PD| is a minimum.

    . . (A)\;12 \qquad (B)\;15 \qquad (C)\;2\sqrt{57} \qquad (D)\;\sqrt{191}\qquad (E)\;\sqrt{241}
    Code:
                                                o D
                                              **
                                            * *
                                          *  *
        A o                             *   *
           * *                        *    * 8
           _*   *                   *     *
         2√3 *     *              *      *
              *       *         *       *
               * 120    *    *   150 *
              B o-----------o---------o C
                :     x   _ P    y    :
                : - - - 3/3 - - - - - :

    \text{Let }\,x \,=\, BP,\;y \,=\, PC\;\text{ where }\,x+y \:=\:3\sqrt{3} \quad\Rightarrow\quad y \:=\:3\sqrt{3}-x


    In \Delta ABP, Law of Cosines:

    . . |AP|^2 \;=\;x^2 + (2\sqrt{3})^2 - 2(x)(2\sqrt{3})\cos120^o \;=\;x^2 + 2\sqrt{3}\,x + 12

    . . |AP| \;=\;\sqrt{x^2 + 2\sqrt{3}\,x + 12}


    In \Delta DCP, Law of Cosines:

    . . |PD|^2 \;=\;8^2 + y^2 - 2(8)(y)\cos150^o \;=\;y^2 + 8\sqrt{3}\,y + 64

    . . |PD| \;=\;\sqrt{y^2 + 8\sqrt{3}\,y + 63} \;=\;\sqrt{(3\sqrt{3}-x)^2 + 8\sqrt{3}(3\sqrt{3}-x) + 64} \;= . \sqrt{x^2-14\sqrt{3}\,x + 163}


    We have: . d \;=\;|AP|+|PD| \;=\;\left(x^2+2\sqrt{3}\,x + 12\right)^{\frac{1}{2}} + \left(x^2 - 14\sqrt{3}\,x + 163\right)^{\frac{1}{2}}

    And that is the function we must minimize.


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


    I'll start it off . . .

    d\:\!' \;\;=\;\;\frac{1}{2}\cdot\frac{2x+2\sqrt{3}}{\sqrt  {x^2+2\sqrt{3}\,x+12}} \:+\: \frac{1}{2}\cdot\frac{2x-14\sqrt{3}}{\sqrt{x^2-14\sqrt{3}\,x+163}} \;\;=\;\;0

    . . . . . \frac{x+\sqrt{3}}{\sqrt{x^2+2\sqrt{3}\,x+12}} \;\;=\;\;-\frac{x-7\sqrt{3}}{\sqrt{x^2-14\sqrt{3}\,x + 163}}


    Square: . \frac{x^2 + 2\sqrt{3}\,x+3}{x^2+2\sqrt{3}\,x+12} \;\;=\;\;\frac{x^2-14\sqrt{3}\,x+147}{x^2-14\sqrt{3}\,x + 163}


    . . and so on . . .



    But check my work . . . please!
    .
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  3. #3
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    Yes your work is right, I found the same solution by thinking D' which is the symmetric of D point with respect to |BC| then making two separate right angled triangles on each side of the question than finding the legs then using the Pythagorean Theorem to find the solution which is \sqrt241
    Thanks anyway how can I mark the thread as solved now.
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