Thread: Applied Optimization Problem

Let Q = (0,4) and R= (9,7) 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.

f(x) = ?

[a,b] a = ? b = ?

The minimal sum of distances is ?


Can anyone show me how to work this, as I am utterly lost. All I can muster is maybe . I'm all fuggled.

There's a little trick that will get the answer quickly. Imagine that the x-axis is a mirror, so that the point R is reflected to the point R'=(9,-7). The shortest distance from Q to R' is a straight line, and you can easily find the point P where that meets the x-axis. Then PQ+PR=PQ+PR', and that will give the solution to the problem.

Originally Posted by pseizure2000

Let Q = (0,4) and R= (9,7) 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.

f(x) = ?

[a,b] a = ? b = ?

The minimal sum of distances is ?


Can anyone show me how to work this, as I am utterly lost. All I can muster is maybe . I'm all fuggled.

PQ = sqrt[(x-0)^2 +(0-4)^2] = sqrt[x^2 +16]
PR = sqrt[(x-9)^2 +(0-7)^2] = sqrt[(x-9)^2 +49]

Let D = PQ +PR
So,
D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***

For maximum or minimum D, dD/dx = 0

dD/dx = [x /sqrt(x^2 +16)] +[(x-9) /sqrt((x-9)^2 +49)]

Set that to zero,

0 = [x /sqrt(x^2 +16)] +[(x-9) /sqrt((x-9)^2 +49)]

Mutiply both sides by the product of the two denominators,

0 = x*sqrt((x-9)^2 +49) +(x-9)sqrt(x^2 +16)

-x*sqrt((x-9)^2 +49) = (x-9)sqrt(x^2 +16)

Square both sides,

(x^2)[(x-9)^2 +49] = [(x-9)^2](x^2 +16)

(x^2)(x-9)^2 +49x^2 = [(x-9)^2](x^2 +16)

49x^2 = [(x-9)^2](x^2 +16) -(x^2)(x-9)^2

49x^2 = [(x-9)^2](x^2 +16 -x^2)

49x^2 = 16(x-9)^2

Take the square roots of both sides,

7x = +,-4(x-9)

7x = 4(x-9)
7x = 4x -36
7x -4x = -36
x = -12 ---------------(i)

7x = -4(x-9)
7x = -4x +36
7x +4x = 36
x = 36/11 = 3.272727.... --------(ii)

If x = -12,
D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***
D1 = sqrt[(-12)^2 +16] +sqrt[(-12 -9)^2 +49]
D1 = sqrt[160] +sqrt[490]
D1 = 34.78 unit long.

If x = 3.272727....,
D = sqrt[x^2 +16] +sqrt[(x-9)^2 +49] ---------------***
D2 = sqrt[(3.272727...)^2 +16] +sqrt[(3.272727.. -9)^2 +49]
D2 = sqrt[26.71] +sqrt[81.80]
D2 = 14.21 unit long.

So, D1 is maximum, and D2 is minimum.

Therefore, for (PQ +PR) to be as small as possible, the P must be at (36/11,0). ---answer.

wow this is great work and 100% correct. thanks, it was quite mind boggling to me.