What is the distance between crossing and factory

[Valentas]

Distance from the sand quarry and brick factory, adjacent to the highway is 30 km and the distance from the sand quarry to the road - 24 km. Building organization has to build a dirt road from the sand quarry to the road. Machines can run for up to 52 km / h on the highway, and the dirt road - up to 20 km / h. How far from the brick factory dirt road has to cross the highway, that the load coming from the sand quarry to the factory would spend the least time?

I believe I should use derivatives and make a function from the drawing. However, I couldn't manage to do so. The answer is 8 km, but I have got different ones. Could anyone make me a function? And explain why it is like you made. Calculations not neccesarry, I'll manage that. Thank you.

 

[soroban]

Hello, Valentas!

This is a classic Calculus problem.
If you've never seen it before, the set-up is quite elusive.

Distance from the sand quarry and brick factory is 30 km;
the distance from the sand quarry to the road is 24 km.
They wish to build a dirt road from the sand quarry to the road.
Machines can run for up to 52 km/h on the highway, and 20 m/h on the dirt road.
How far from the brick factory should the dirt road cross the highway
so that the load coming from the sand quarry to the factory would spend the least time?

I revised the diagram.

    Q o
      |* *
      | *   *
      |  *     *   30
   24 |   *       *
      |    *         *
      |     *           *
      |      *             *
    R o-------o---------------o F
      : 18-x  P - - - x - - - :
      : - - - - - 18  - - - - :

The sand quarry is at \(\displaystyle Q.\)
The brick factory is at \(\displaystyle F\!:\;QF = 30\)
The road is \(\displaystyle RF\!:\;QR = 24\)
Pythagorus says: \(\displaystyle RF = 18\)

The dirt road runs from \(\displaystyle Q\) to \(\displaystyle P.\)
Let \(\displaystyle x = PF\text{, then }RP = 18-x\)

The dirt road has length \(\displaystyle QP.\)
In right triangle \(\displaystyle QRP\!:\;QP^2 \:=\18-x)^2 + 24^2 \:=\:x^2-36x + 900 \)


They will drive \(\displaystyle \sqrt{x^2-36x+900}\) km at 20 km/h.

. . This will take:.\(\displaystyle \dfrac{\sqrt{x^2-36x+900}}{20}\text{ hours.}\)


Then they will drive \(\displaystyle x\) km on the highway at 52 km/h.

. . This will take:.\(\displaystyle \dfrac{x}{52}\,\text{ hours.}\)

Hence, the total time is: .\(\displaystyle T \;=\;\dfrac{(x^2-36x+900)^{\frac{1}{2}}}{20} + \dfrac{x}{52}\text{ hours}\)

And that is the function we must minimize.

 

[Valentas]

This is amazing explanation! Thank you very much. I had already went through the problem till you get that square root. Thank you again. Now I will learn and know how to solve this type of problem for myself