Stevie completes a quest by travelling from #A# to #C# vi #P#. The speed along #AP# is 4 km/hour, and along #AB# it is 5 km/hour. Solve the following?

A) The distance #D(x)# from A to C via P as a function of #x#.

By Pythagoras;

# \ \ \ \ \ AP^2 = AB^2 + BP^2 #
# :. AP^2 = 3^2 + x^2 #
# :. AP^2 = 9 + x^2 #
# :. \ \ AP = sqrt(9 + x^2) # (must be the +ve root)

Then;

# \ \ \ \ \ D(x) = AP + PC #
# :. D(x) = sqrt(9 + x^2) + (6-x) #

B) Find the time, #T(x)#, that is required to travel from A to C via P.

Using # "speed" = "distance"/"time" #:

Along AP the speed is #4# km/hour, provided #x>0# (otherwise we are going along AB at #5# km/hour) and so:

# " "4 = sqrt(9 + x^2)/t_1 #
# :. t_1 = sqrt(9 + x^2)/4 #

Along PC (or AB) the speed is #5# km/hour, and so:

# " " 5 = (6-x)/t_2 #
# :. t_2 = (6-x)/5 #

And so, the total time is given by:

# \ \ \ \ \ T(x) = t_1 + t_2 #
# :. T(x) = sqrt(9 + x^2)/4 + (6-x)/5 # for (#x gt 0#)

C) Stevie's Quest

In order for Stevie to complete the quest we need to find a critical point of T(x):

Differentiating wrt #x# we get:

# " "T'(x) = 1/4*1/2(9+x^2)^(-1/2)*2x+1/5(-1) #
# :. T'(x) = x/(4sqrt((9+x^2))) - 1/5 #

At a critical point, #T'(x)=0#

# => x/(4sqrt(9+x^2)) - 1/5 = 0 #
# :. 5x - 4sqrt(9+x^2) = 0 #
# :. 5x = 4sqrt(9+x^2) #
# :. 25x^2 = 16(9+x^2) #
# :. 25x^2 = 144 + 16x^2 #
# :. 9x^2 = 144 #
# :. x^2 = 144/9 #
# :. x^2 = 16 #
# :. x = 4 # (must be the +ve root)

When #x=4# we have:

# :. D(4) = sqrt(9 + 16) + (6-4) = 7#
# :. T(4) = sqrt(9 + 16)/4 + (6-4)/5 = 33/20 = 1.65#

We can confirm visually that this corresponds to a minimum by looking at the graph:

graph{sqrt(9 + x^2)/4 + (6-x)/5 [-15, 15, -1, 10]}