[ASK] Minimum Length of AP + PB

[Monox D. I-Fly]

The point A is located on the coordinate (0, 5) and B is located on (10, 0). Point P(x, 0) is located on the line segment OB with O(0, 0). The coordinate of P so that the length AP + PB minimum is ....

A. (3, 0)

B. (3 1/4, 0)

C. (3 3/4, 0)

D. (4 1/2, 0)

E. (5, 0)


What I did:

f(x) = AP + PB =[MATH]\sqrt{5^2+x^2}+(10-x)=\sqrt{25+x^2}+10-x[/MATH]

In order to make AP + PB minimum, so:

f'(x) = 0

[tex]\frac12(25+x^2)^{-\frac12}(2x)+(-1)=0[/tex]

[tex]\frac{x}{\sqrt{25+x^2}}=1[/tex]

[tex]x=\sqrt{25+x^2}[/tex]

[tex]x^2=25+x^2[/tex]

This is where I got stuck. Subtracting [tex]x^2[/tex] from both sides would leave me with 0 = 25 which is obviously incorrect. Where did I do wrong?

[HallsofIvy]

Are you sure you have stated the problem correctly? Since the shortest distance between two points is a straight line, the minimum value for AP+ PB is AB when P= B. But (10, 0) is not given as a possible answer.

The reason you ran into a problem is that a differentiable function always has a minimum on a closed and bounded interval either where the derivative is 0 or at an endpoint. Here, the derivative is never 0 so the minimum is at the endpoint, P= B.