Question

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.

Answer 1

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

Answer 2

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

Answer 3

Who is Goku...

Answer 4

since this is not related to question, please continue in private messages/chat...

Answer 5

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)

Answer 6

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 ??

Answer 7

It does, so I have everything figured out except how you find what b=

Answer 8

never mind I got it
Thanks for the help!

Answer 9

good :) ask if any further doubts...