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Varku

Varku

Space Marshal
Apr 21, 2016
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Varku
Ezz said:
Take that A+ away. Looking at it today I made a dumb mistake.

I took the diagonal of the box (sqrt(1² + 1²)) as hc but that's wrong since it's not orthogonal to the ladder. (That would only be the case if the box is ~ 2,12m x 2,12m )
Maybe I'll give it another try sometime
Click to expand...
okay, I've got my sheet of paper from work, and here is my solution to the propblem:

Code: 
b|
.|\
.|..\
1|XXX.\P
.|XXX...\
______________
.    1  c
P needs to be a point in the equation so this must be true:
P=(1|1) => a*1+b=1 | -b
a*1=1-b
a=1-b

we want the ladder to touch the ground, so this must also be true:
f(c)=0 => a*c+b=0 |-b
a*c=-b | :a
c=-b/a

thanks to pythagoras we can say:
b²+c²=3²

with the previus functions we can replace c with -b/a
and we can replace a with 1-b
so we get this lovely function:

b²+(-b/[1-b])²=9
sadly it's not in a "nice" form, so we do some "magic"
b²+([-b]²/[1-b]²)=9 |*(1-b)²
b²*(1-b)² +(-b)²=9*(1-b)²
b²(b²-2b²+1)+b²=9(b²-2b²+1)
b⁴-2b³+b²+b²=9b²-18b²+9 |-(9b²-18b²+9)
b⁴-2b³-7b²+18b-9=0
finaly a "nice" form

solve for b, I've used this site to compute it:
https://www.wolframalpha.com/input/?i=b⁴-2b³-7b²+18b-9=0

there are 4 solutions for this function, but as we're looking for a maximum for b, we can choose the biggest number.

b= 1/2 (1+sqrt(10)+sqrt(7-2*sqrt(10)) ~ 2,49207

let's insert that into our other equation to validate the solution.

a=1-b
=> a=1-2,49207
=> a=-1,49207

c=-b/a
=>c=-2,49207/-1,49207
=>c=1,67021
as this is equal to an other solution for b⁴-2b³-7b²+18b-9=0 our solutions seems to be right.

tada, now we know, that the ladder can reach as high as ~2.49207m.
 
Last edited:
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Sraika

Sraika

Space Marshal
Nov 7, 2017
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Sraika
Varku said:
okay, I've got my sheet of paper from work, and here is my solution to the propblem:

Code: 
b|
.|\
.|..\
1|XXX.\P
.|XXX...\
______________
.    1  c
P needs to be a point in the equation so this must be true:
P=(1|1) => a*1+b=1 | -b
a*1=1-b
a=1-b

we want the ladder to touch the ground, so this must also be true:
f(c)=0 => a*c+b=0 |-b
a*c=-b | :a
c=-b/a

thanks to pythagoras we can say:
b²+c²=3²

with the previus functions we can replace c with -b/a
and we can replace a with 1-b
so we get this lovely function:

b²+(-b/[1-b])²=9
sadly it's not in a "nice" form, so we do some "magic"
b²+([-b]²/[1-b]²)=9 |*(1-b)²
b²*(1-b)² +(-b)²=9*(1-b)²
b²(b²-2b²+1)+b²=9(b²-2b²+1)
b⁴-2b³+b²+b²=9b²-18b²+9 |-(9b²-18b²+9)
b⁴-2b³-7b²+18b-9=0
finaly a "nice" form

solve for b, I've used this site to compute it:
https://www.wolframalpha.com/input/?i=b⁴-2b³-7b²+18b-9=0

there are 4 solutions for this function, but as we're looking for a maximum for b, we can choose the biggest number.

b= 1/2 (1+sqrt(10)+sqrt(7-2*sqrt(10)) ~ 2,49207

let's insert that into our other equation to validate the solution.

a=1-b
=> a=1-2,49207
=> a=-1,49207

c=-b/a
=>c=-2,49207/-1,49207
=>c=1,67021
as this is equal to an other solution for b⁴-2b³-7b²+18b-9=0 our solutions seems to be right.

tada, now we know, that the ladder can reach as high as ~2.49207m.
Click to expand...
Just move the damn box.
 
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Irishnutcase

Irishnutcase

Space Marshal
Jan 27, 2015
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Irishnutcase
#tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow #tablethrow
 
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Ezz

Ezz

Space Marshal
Feb 4, 2016
943
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Pfen
Varku said:
okay, I've got my sheet of paper from work, and here is my solution to the propblem:

Code: 
b|
.|\
.|..\
1|XXX.\P
.|XXX...\
______________
.    1  c
P needs to be a point in the equation so this must be true:
P=(1|1) => a*1+b=1 | -b
a*1=1-b
a=1-b

we want the ladder to touch the ground, so this must also be true:
f(c)=0 => a*c+b=0 |-b
a*c=-b | :a
c=-b/a

thanks to pythagoras we can say:
b²+c²=3²

with the previus functions we can replace c with -b/a
and we can replace a with 1-b
so we get this lovely function:

b²+(-b/[1-b])²=9
sadly it's not in a "nice" form, so we do some "magic"
b²+([-b]²/[1-b]²)=9 |*(1-b)²
b²*(1-b)² +(-b)²=9*(1-b)²
b²(b²-2b²+1)+b²=9(b²-2b²+1)
b⁴-2b³+b²+b²=9b²-18b²+9 |-(9b²-18b²+9)
b⁴-2b³-7b²+18b-9=0
finaly a "nice" form

solve for b, I've used this site to compute it:
https://www.wolframalpha.com/input/?i=b⁴-2b³-7b²+18b-9=0

there are 4 solutions for this function, but as we're looking for a maximum for b, we can choose the biggest number.

b= 1/2 (1+sqrt(10)+sqrt(7-2*sqrt(10)) ~ 2,49207

let's insert that into our other equation to validate the solution.

a=1-b
=> a=1-2,49207
=> a=-1,49207

c=-b/a
=>c=-2,49207/-1,49207
=>c=1,67021
as this is equal to an other solution for b⁴-2b³-7b²+18b-9=0 our solutions seems to be right.

tada, now we know, that the ladder can reach as high as ~2.49207m.
Click to expand...
Okay that was actually not that complicated ^^'
But thanks for demonstrating :ok_hand:
 
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