Calculas problem, I need help!

At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 17 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 5 PM?
(Note: 1 knot is a speed of 1 nautical mile per hour.)
I have tried this problem many times and i get the answer as 26.2488, BUT IT IS WRONG, I don't know what i am doing wrong. Please help
Thanks!

let B's position at noon be (0,0), then A's position is (-30-17t,0)
B's position is (0,20t)
distance between A and B are
D = sqrt((30+17t)^2 + (20t)^2) = sqrt(689t^2 + 1020t+900)
dD/dt = 1/2(689t^2+1020t+900)^(-1/2) * (2*689t+1020)
substitute t = 5
dD/dt = 25.95 knots/hr

At noon, imagine ship B is at the origin and ship A is 30 miles along the negative x-axis. After 5 hours (5pm) ship A has sailed 30 + (17 X 5)

= 115 miles West

In this time ship B has sailed 20 X 5 = 100 miles North. Construct a right angled triangle of horizontal length x = 115 mi, vertical length y = 100 mi. Using Pythagoras, the distance between A and B is given by

r² = x² + y² = 115² + 100² = 23225

r = √23225
r = 152 miles.

Now, using x² + y² = r² we differentiate each side with respect to time.

2x(dx/dt) + 2y(dy/dt) = 2r(dr/dt)

Or x.dx/dt + y.dy/dt = r.dr/dt. The quantity we want to find is dr/dt which is the rate the distance between the ships is increasing.

Speed of ship A is 17 knots = dx/dt
Speed of B is 20 knots = dy/dt

x = 115 , y = 100 and r = 152

(152)dr/dt = (115)(17) + (100)(20)

= 3955

dr/dt = 3955/152

dr/dt = 26 knots which is much about the same as your answer.