AHSME Challenge Math Question Cowboy A cowboy is $4$ miles south of a stream which flows due east. He is also $8$ miles west and $7$ miles north of his cabin. He wishes to water his horse at the stream and return home. The shortest distance (in miles) he can travel and accomplish this is:
a. $4 + \sqrt{185}$
b. $16  $
c. $17  $
d. $18  $
e. $\sqrt{32} + \sqrt{137}$
The correct answer is C, however I did not get that. I got my incorrect answer (A) by adding $7$ and $4$. I used the Pythagorean Theorem to find the hypotenuse length of an $8\times 11$ triangle which is $\sqrt{185}$.
Could you please explain how to get it? Thank you.

Worth mentioning is the 'honest' way to do the problem: find the value $x$, the east distance the cowboy travels, such that the total distance $T(x)=\sqrt{x^2+16}+\sqrt{(8-x)^2+11^2}$ is minimized. Differentiating, we have
$$
T'(x) = \frac{x}{\sqrt{x^2+16}} +\frac{x-8}{\sqrt{x^2-16x+185}}
$$Set this equal to zero and solve for $x$:
$$
0= x\sqrt{x^2-16x+185}+(x-8)\sqrt{x^2+16}
$$
$$
(x-8)^2(x^2+16)= x^2(x^2-16x+185)
$$
$$
105 x^2+256x-1024=0
$$Solving this gives $x=-32/7$, which can obviously be discarded, and $x=32/15$; this value gives $D_{\text{to stream}}=68/15$ and $D_{\text{to cabin}}=187/15$, for a total distance of $17$.
