Stuck with an old math question

[SKG_Scintill]

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So basically I got bored during math class (if you can call this math) and I decided to redo a question of my math exam on high school.

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The question was something like this:

There are two buildings along a road. The distances between the buildings and the road are 50 meters and 80 meters respectively. The distance between the two buildings is 100 meters (from the center of the building, don't take the dimensions of the building into consideration).

They want to build a bus stop on the road and make two straight walkways from the bus stop to the buildings (two Pythagorean triangles basically).
At what distance is the length of the two walkways combined the smallest?

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What I have is this:

f(x) = sqrt(50^2 + x^2) + sqrt(80^2 + (100-x)^2)
I hope this formula makes sense. It's the length of the two hypoteni combined.

f(x) = sqrt(x^2 + 2500) + sqrt(x^2 - 200x + 16400)
f(x) = (x^2 + 2500)^(1/2) + (x^2 - 200x + 16400)^(1/2)
say u = x^2 + 2500 say v = x^2 - 200x + 16400
f(x) = u^(1/2) + v^(1/2)
f'(x) = u(-1/2) * u' + v(-1/2) * v' (chain rule)
f'(x) = 1/(2*sqrt(x^2 + 2500)) * 2x + 1/(2*sqrt(x^2 - 200x + 16400)) * (2x - 200)
f'(x) = 2x/(2*sqrt(x^2 + 2500)) + (2x - 200)/(2*sqrt(x^2 - 200x + 16400))

2x/(2*sqrt(x^2 + 2500)) + (2x - 200)/(2*sqrt(x^2 - 200x + 16400)) = 0

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This is where I got stuck. I tried doing the square of it and then dividing everything by 4.

2x/(2*sqrt(x^2 + 2500)) + (2x - 200)/(2*sqrt(x^2 - 200x + 16400)) = 0
4x^2/(4(x^2 + 2500)) + (4x^2 - 800x + 40000)/(4*(x^2 - 200x + 16400)) = 0
4x^2/(4x^2 + 10000) + (4x^2 - 800x + 40000)/(4x^2 - 800x + 65600) = 0
x^2/(x^2 + 2500) + (x^2 - 200x + 10000)/(x^2 - 200x + 16400) = 0

That just made the numbers different, solving it was still the question.

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I hope it's clear. When typing was it, it felt quite hectic...
Maybe someone can help?

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[dag12]

Re: Stuck with an old math question

starting from 2x/(2*sqrt(x^2 + 2500)) + (2x - 200)/(2*sqrt(x^2 - 200x + 16400)) = 0,
move one term over to the other side of the equal sign. Then cross multiply. You should be able to solve it from there.

[emerald000]

Re: Stuck with an old math question

Let's see:

The distance between the buildings at the road is sqrt(100²-(80-50)²) = 10 sqrt(91).



So you will want to minimize y = sqrt(50²+x²) + sqrt(80²+(10sqrt(91)-x)²)

You can follow dag12's advice once you get to the same step.

[igotrhythm]

Re: Stuck with an old math question

This is where your calculus comes in. To find a local minimum/maximum, take the derivative of the right side with respect to x (there's a lot of chain-ruling going on here, so be careful) then define 0 <= x <= 10sqrt(91), x being the distance from one arbitrarily selected building to the bus stop. Find all values of x for which the derivative equals zero, and those will be your critical numbers.

Go back to the original expression Emerald posted up, substitute each critical number for x as well as the endpoints of the interval to see which value of x yields the smallest value for y. This will be your answer.

[SKG_Scintill]

Re: Stuck with an old math question

Dammit, I knew it would be unclear somewhere.
The length that is 100 is along the road, where 10*sqrt(91) is now

Also I figured it out with dag12's advice, should've mentioned that earlier