Question 1153729
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Introduce the coordinate plane (x,y) in a way that ship's A initial location is the origin of the coordinate plane;
x-axis is directed East and y-axis is directed North.


Then the initial location of the ship B is the point ({{{-4/sqrt(2)}}},{{{4/sqrt(2)}}}).


The ship A moves along y-axis, according to equation y = 6*t, where t is the time in hours.

So, the ship A location in time on the coordinate plane is A = (0,6t).


Ship B moves parallel to x-axis, according to equation x = {{{-4/sqrt(2)}}} + 4*t, where t is the time in hours.

So, the ship B location in time on the coordinate plane is B = ({{{-4/sqrt(2)+4t}}},{{{4/sqrt(2)}}}).


Now find the square of distance between the points A and B


    d^2 = {{{(-4/sqrt(2)+4t)^2}}} + {{{(4/sqrt(2)-6t)^2}}}.


Simplify


    d^2 = {{{16/2 - (sqrt(2)/2)*2*4*4t + 16t^2}}} + {{{16/2 - (sqrt(2)/2)*2*4*6t + 36t^2}}} = {{{16 - sqrt(2)*40t + 52t^2}}}.


Thus you have this quadratic function of "t", and you need to find its MINIMUM.


Use the formulas for the vetex of the quadratic function/parabola


    {{{t[min]}}} = " (-b/(2a)) " = {{{(40*sqrt(2))/(2*52)}}} = {{{(10*sqrt(2))/26}}} = {{{(5*sqrt(2))/13}}} = 0.5439 hours.


To find the square of the minimal distance, substitute t = 0.5439 into the formula (1). You will get


    d^2 = {{{16 - sqrt(2)*40*0.5439 + 52*0.5439^2}}} = 0.6154.


Hence d = {{{sqrt(0.6154)}}} = 0.7845.


<U>ANSWER</U>.  The minimal distance between the ships will be 0.7845 kilometers.
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Notice that I did not use Calculus in my solution.

Algebra is just enough.


But if you want to use &nbsp;Calculus &nbsp;to find &nbsp;{{{t[min]}}}, &nbsp;you can do it.


The route &nbsp;(the road) &nbsp;is open for you now.