What is the Minimum Length of AP + PB?

[Monoxdifly]

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 =$$\sqrt{5^2+x^2}+(10-x)=\sqrt{25+x^2}+10-x$$
In order to make AP + PB minimum, so:
f'(x) = 0
$$\frac12(25+x^2)^{-\frac12}(2x)+(-1)=0$$
$$\frac{x}{\sqrt{25+x^2}}=1$$
$$x=\sqrt{25+x^2}$$
$$x^2=25+x^2$$
This is where I got stuck. Subtracting $$x^2$$ from both sides would leave me with 0 = 25 which is obviously incorrect. Where did I do wrong?

 

[skeeter]

uhh ...

$\dfrac{x}{\sqrt{x^2+25}} < 1$ for any value of $x$

 

[Monoxdifly]

So, which steps should I fix? And become what?

 

[Opalg]

There is something wrong with this question. The length AP + PB will be a minimum when APB is a straight line, and that happens when P = B. So the answer should be that P has coordinates (10,0).

Notice that to allow for the possibility $x>10$, the formula for $f(x)$ should be $\sqrt{5^2+x^2} + |10-x|$, which is indeed minimised at $x=10$.

 

[Monoxdifly]

There is something wrong with this question. The length AP + PB will be a minimum when APB is a straight line, and that happens when P = B. So the answer should be that P has coordinates (10,0).

Come to think of it, you're right. APB is a triangle thus the minimum length of AP + PB should be AB. Thanks. Glad I'm not the one who messed up.

 

[skeeter]

APB is a triangle ...

The problem statement does not say APB is a triangle. I believe the problem's aim was to show the possibility of an endpoint minimum.

 

[Monoxdifly]

I think I should have said that APB is "supposed to be" a triangle, not necessarily a triangle itself.