## heradog Group Title 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. one year ago one year ago

1. hartnn

have you tried this ? do you know distance formula to get PQ and PR ?

2. yololol

YOU'RE NOT EVEN GOKU HOW DARE YOU USE HIS PICTURE HE WAS MY ONLY FRIEND AS A CHILD

Who is Goku...

4. hartnn

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)

6. hartnn

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!

9. hartnn

good :) ask if any further doubts...