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]}