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A point $ P $ needs to be located somewhere on the line $ AD $ so that the total length $ L $ of cables linking $ P $ to the points $ A $, $ B $, and $ C $ is minimized (see the figure). Express $ L $ as a function of $ x = | AP | $ and use graphs of $ L $ and $ dL/dx $ to estimate the minimum value of $ L $.

$$9.35 \mathrm{m}$$

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All right. So you're asked to find a value of X. That will minimize this distance. L which is the length of cable A PPB pc. And we're told to let X equal ap So that means P. D. Would be five minus X. Because the whole thing's five take away that we'll give you five minus axe. Now that gives me enough information that I can now find PB So this is a right triangle. So using the pythagorean theorem, P B squared will equal P. D squared plus B. D squared. So the absolute value would be the square root of five minus X squared plus four because two squared is four. And then using the same process on the other side, P C squared would be P. D squared plus C. D squared. So the absolute value of pc would be the square root of five minus X squared plus nine. And then I like to simplify this when I type it into the computer for graphing. So that's 25 minus 10 X plus X squared plus four. And or 24 25 sorry. And 25 plus four is 29. So go ahead and make that a 29. Then over here this will square out to be the same thing as 25 minus 10 X plus X squared. But now I need to add nine to it And nine plus 25 will give 34. And then when I put it all together, need a little more room here. When I put them all, the other L. Of X. Is X plus the square root of X. Word minus 10 X plus 29 plus. I didn't give myself enough room. The square root of X squared minus 10 X plus 34. So I'm gonna take that function and graph it on a graphing utility. I like to use demos dot com because it is free. So I've already got it opened up here and graft. I just need to make my screen a little bigger. So I used Carrot 1/2, but it's uh I could have used the square root, so I've got it all typed in here. So the red function is my graph. Now. I could stop there because automatically that gives me the minimum value the way that programs designed. But since it asked me to use the graphs to estimate the minimum value to do the first derivative, it's D. D. X. And then brackets L. Of X. And here's my first derivative. And remember that value The minimum occurs were the first derivative equals zero. So you can see the X. Value that I would need Would be 3.593. And then the minimum value of L would be 9.352.

Mount Vernon Nazarene University