The Minimum Value of The Sum

• May 1st 2010, 11:58 AM
JohnDoe
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 $\sqrt241$
• May 1st 2010, 08:25 PM
Soroban
Hello, JohnDoe!

It can be solved without a calculator,
. . but it takes a lot of Algebra . . .

Quote:

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!
.
• May 2nd 2010, 09:50 AM
JohnDoe
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.