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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.

Calculus1
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have you tried this ? do you know distance formula to get PQ and PR ?
YOU'RE NOT EVEN GOKU HOW DARE YOU USE HIS PICTURE HE WAS MY ONLY FRIEND AS A CHILD
Who is Goku...

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since this is not related to question, please continue in private messages/chat...
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)
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 ??
It does, so I have everything figured out except how you find what b=
never mind I got it Thanks for the help!
good :) ask if any further doubts...

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