My Math Forum Problem involving Chain Rule, Finding Maximum/Minimum point

 Calculus Calculus Math Forum

 February 5th, 2009, 06:39 AM #1 Newbie   Joined: Feb 2009 Posts: 1 Thanks: 0 Problem involving Chain Rule, Finding Maximum/Minimum point [color=#400080]here's the problem: Two points, A and B are located 2km and 3km respectively, opposite from a shoreline. What point C on the shoreline could be chosen for Spot X to minimize the sum of the distances from A and B to the Spot X, if the distance between D and E is 4km? the book says the answer is 8/5. Please show the solution.[/color] Illus. in this link: http://i172.photobucket.com/albums/w33/ ... _14/ll.jpg
 February 5th, 2009, 04:54 PM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: Problem involving Chain Rule, Finding Maximum/Minimum point The answer can be found by using the fact that the minimum of a differentiable function, if it is not located at a boundary point, is a point at which the tangent line is horizontal, i.e., $\frac{dy}{dx}\,=\,0$. Using the Pythagorean theorem, we can see that the distances from $A$ and $C$ to $B$ are $\begin{eqnarray*} s_1 &=& \sqrt{x^2\,+\,4\,\mbox{km}^2} \\ s_2 &=& \sqrt{(4\,\mbox{km}\,-\,x)^2\,+\,9\,\mbox{km}^2}\,=\,\sqrt{x^2\,-\,8x\,\mbox{km}\,+\,25\,\mbox{km}^2}. \end{eqnarray*}$ The minimum value of the sum of the two distances, if it is not located at a boundary point, will be at a point where $\frac{ds}{dx}\,=\,\frac{d}{dx}(\sqrt{x^2\,+\,4\,\m box{km}^2}\,+\,\sqrt{x^2\,-\,8x\,\mbox{km}\,+\,25\,\mbox{km}^2})\,=\,0.$
 February 6th, 2009, 09:34 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 Let FADG be a straight line such that FD = 3km and DG = 2km, then AB + BC = GB + BC, which is minimized when CBG is a straight line, in which case BD/DG = CF/FG, so x = BD = DG(CF/FG) = 2km(4km/5km) = 1.6km.

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### maximum and minimum values involving chain rule

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