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heradog

  • 3 years ago

Let Q = (0,4) and R= (12,8) be given points in the plane. We want to find the point P=(x,0) on the x-axis such that the sum of distances PQ+PR is as small as possible. (Before proceeding with this problem, draw a picture!) To solve this problem, we need to minimize the following function of x: f(x)= over the closed interval [a,b] where a= and b=. We find that f(x) has only one critical point in the interval at x= where f(x) has value Since this is smaller than the values of f(x) at the two endpoints, we conclude that this is the minimal sum of distances.

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  1. hartnn
    • 3 years ago
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    have you tried this ? do you know distance formula to get PQ and PR ?

  2. yololol
    • 3 years ago
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    YOU'RE NOT EVEN GOKU HOW DARE YOU USE HIS PICTURE HE WAS MY ONLY FRIEND AS A CHILD

  3. rebeccaskell94
    • 3 years ago
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    Who is Goku...

  4. hartnn
    • 3 years ago
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    since this is not related to question, please continue in private messages/chat...

  5. heradog
    • 3 years ago
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    I found where a similar problem was worked, but didn't quite understand what their numbers meant PQ = √[(x-0)2 + (0-6)2] = √(x2 + 36) PR = √[(x-6)2 + (0-7)2] = √(x2 - 12x + 85) f(x) = √(x2 + 36) + √(x2 - 12x + 85)

  6. hartnn
    • 3 years ago
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    Distance between points (x1,y1) and (x2,y2) is \(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\) does this help ??

  7. heradog
    • 3 years ago
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    It does, so I have everything figured out except how you find what b=

  8. heradog
    • 3 years ago
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    never mind I got it Thanks for the help!

  9. hartnn
    • 3 years ago
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    good :) ask if any further doubts...

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