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