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Hey people. It's the first time I'm asking this (whoops ok second one  ), but it's been 2 days since I've been trying this problem which should be easy as hell for me (it's just simple calculus) and it's driving me crazy.
If someone could shed some light on it, it would be lovely.
"A fence is 27/8 meters tall, and there is a house 8 meters away from it, find the shortest possible stair that goes from the ground, over the fence and touches the house."
Here, I made a drawing of it:
+ Show Spoiler +
This is simply about finding a function (for the stair length (L)) and then diferentiating it and finding its minimum point (by equaling the derivative to 0 and solving the equation, etc.)
The problem is that I have tried to postulate the function in half a dozen ways now and the derivative (equaled to 0) is never an equation below grade 4, and that just seems too complicated.
I have tried using Θ, like cosΘ = (x+8)/L, then cosΘ = x/sqrt(x²+8²). And then just substituting.
Or like sinΘ= 27/(8x) and then squaring both the cosine and sine and summing them, etc.
I've tried using H instead of Θ too and and well as I have said, I haven't found an equation below 4th grade.
So if anyone could help it would be greatly appreciated.
By the way, the answer is supposed to be 125/8 meters if that helps.
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Apologies if you've tried this and I'm just bad at reading formulas, but:
Have you tried breaking it up into a sum of 2 triangles that have the same angles? One with base x, another with base 8 meters.
*edit* Synapse>> Yoink. 
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So here is my thought, though it didn't work out for me.
L^2=(x+8)^2 + H^2
tan(theta)=27/8x from the little triangle
tan(theta)=H/(x+8) from the big triangle
So we can equate these:
27/8x=H/(x+8)
Solving for H:
H=27(x+8)/8x
Then we can sub into our equation for L:
L^2=(x+8)^2+ (27(x+8)/8x)^2
Then take the derivative w/r to x and gogogo.
it didn't work out for me, but I'll give it another shot. I don't see my error yet.
EDIT: need L^2 heh, i'll try it now
not working...whyyy?
Ah so i used that program the guy above me used...coool
http://www.wolframalpha.com/input/?i=d/dx(sqrt(x^2+16x+64+(729/64)+(729/(4x))+(729/(x^2))))
I get a root in the derivative at 9/2
Evaluating at my function gives me 125/8.
I imagine that finding that root by hand would be tough.
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On December 04 2009 09:46 synapse wrote:apparently the differentiated function I wrote would look something like
Yeah that ends up being something like a 10th grade equation 
On December 04 2009 09:55 Zortch wrote:
So here is my thought, though it didn't work out for me.
L^2=(x+8)^2 + H^2
tan(theta)=27/8x from the little triangle
tan(theta)=H/(x+8) from the big triangle
So we can equate these:
27/8x=H/(x+8)
Solving for H:
H=27(x+8)/8x
Then we can sub into our equation for L:
L^2=(x+8)^2+ (27(x+8)/8x)^2
Then take the derivative w/r to x and gogogo.
it didn't work out for me, but I'll give it another shot. I don't see my error yet.
EDIT: need L^2 heh, i'll try it now
I hadn't tried using both theta and H before, but your function seems rather big as well ~.~
Maybe there's no hope but to solve a quartic equation. Which sounds ridiculous for such a simple exercise.
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Zortch's way works well. It's easiest to minimize L^2 (instead of taking square roots and minimizing L) -- that way there aren't any square roots to deal with when you differentiate. You ought find
d(L^2)/dx = 2(x+8)(1 - 27^2/(8x^3));
setting this equal to zero makes solving for x easy -- no quartic equation, just x^3 = 27^2/8.
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I had to use Mathematica, but I was able to get the correct answer.
Basically you split the area under the ladder into two triangles and a rectangle. One of the triangles has base x and height 27/8, while the other has base 8 and unknown height H. Since the 2 triangles are similar, we do 27/(8x) = H/8 to get that H = 27/x.
Now we take our two triangles and write L in terms of x:
L = sqrt(x^2 + (27/8)^2) + sqrt(8^2 + (27/x)^2)
I took the derivative of this using Mathematica and it gave me a root at x=9/2.
Plug this back into the equation for L to get that L=125/8.
Hope this helps, I still have no idea how to do it by hand.
Edit: Seems like people above me got the same thing
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Oh thanks guys, that's pretty cool. I thought of minimizing the square but that just seemed weird to me.
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I think it would be easiest to solve for the angle instead.
We have 27/(8*tan(v))=x and (x+8)/cos(v)=y, together we get (after some rearrangement) 8/cos(v) + 27/(8*sin(v)) = y, if we derive we get y' = - 8sin(v)/cos^2(v) + 27cos(v)/(8*sin^2(v), for extreme value we let y' be 0 and get 64sin^3(v) = 27cos^3(v) => 4sin(v) = 3cos(v) => arctan(3/4) = v.
Edit. Just to finish my calculations, we have
27/(8*0.75) = x = 4.5,
8 + 4.5 = 12.5
12.5/cos(arctan(3/4)) = 12.5/0.8 = 15.625 = 125/8
so that is the analytical solution.
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Pretty cool solution as well Fwmeh thanks.
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United States928 Posts
On December 04 2009 10:21 Cloud wrote:
Oh thanks guys, that's pretty cool. I thought of minimizing the square but that just seemed weird to me.
![[image loading]](http://img223.imageshack.us/img223/3817/kruemelmonsteryn0.gif)
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Sorry for bumping this thread guys but I wanted to just give it some more closure (actually...to give myself closure  ).
I was reading this thread and thought the approach of minimising the square was a smart one. It intuitively makes sense that if the original function produces only positive values, then the squared function should have the minimum point at the same x value.
What bugged me was how to prove this mathematically for a general positive function. I decided to play around with it a bit on paper and scribble some things because I like doing little math problems. I came up with this 'proof' (it's probably not very proper but I only have high school level math under my belt so I'm noob ) and it surprised me how simple it was actually:
Let y=f(x) and let f(x) > 0
dy/dx = f'(x)
To find minimum point, let dy/dx = 0 which means f'(x) = 0
--> solve and get x = a where "a" is the x value of the minimum point on the original curve
Now y^2 = f(x) x f(x)
d(y^2)/dx = f'(x)f(x) + f(x)f'(x) (product rule) = 2f'(x)f(x)
Let d(y^2)/dx = 0 which means 2f'(x)f(x) = 0
--> either:
1. f(x) = 0 --> no real solutions because f(x) > 0 or
2. f'(x) = 0 --> x = a which is the same solution we get by differentiating the original function. Woo!
Probably noone cares but now I can sleep easy tonight
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EPT![[image loading]](http://i50.tinypic.com/kdlfsk.jpg)
