That is a good start and I got the same.
Now just like guest said you take the first derivative and set it equal to 0.
When you get the answer I suppose you should check that it give a the time when the distance is minimum but just looking at the problem this will obviouly be the case.
I am also going to change your d into an x,
\(x=\sqrt{(150t)^2+(15-400t)^2}\\ x=[(150t)^2+(15-400t)^2]^{0.5}\\ \frac{dx}{dt}=0.5[(150t)^2+(15-400t)^2]^{-0.5}[300t*150+2(15-400t)*-400]\\ \frac{dx}{dt}=\frac{0.5[300t*150+2(15-400t)*-400]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=\frac{0.5[45000t+800(400t-15)]}{ [(150t)^2+(15-400t)^2]^{0.5} }\\ \frac{dx}{dt}=0\;\;when\\ 0.5[45000t+800(400t-15)]=0\\ 450t+8(400t-15)=0\\ 3650t-120=0\\ t=\frac{12}{365}\;\;hours\\ When \;\;t=\frac{12}{365}\;\;hours\\ x=[(150*\frac{12}{365})^2+(15-400*\frac{12}{365})^2]^{0.5}\\\)
((150*12/365)^2+(15-400*12/365)^2)^0.5 = 5.266851623825875 = 5.27 miles
There you go. My answer is exactly the same as guests!
Good work guest. Can 2 'great' minds both be wrong ?