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Who did BOX + 2 Triangles + p + q question? (1 Viewer)

Who did the QUESTION with the BOX and p, g, triangles

  • YES

    Votes: 9 27.3%
  • NO

    Votes: 24 72.7%

  • Total voters
    33

VanCarBus

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Huh?
I got a full on blackout on this question. Anyone else?
 

symo

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i could find the first angle but thats as far as i got
 

Constip8edSkunk

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use the fact that the diagonal divide the square into 2 isosceles triangles, so beta=arctan1/q
use triangle angle sum, add them together, tan both sides
area=1-area of 2 triangles
after that differentiate to get the value for p that maximise area, sub it back in.
 

felix_js

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hmm i got 1/2.....took limit as p goes to 0 and it comes out as half
 
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ND

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Yep, 1/2 is right... i dunno wtf i was doing... i hate careless mistakes....
 

Constip8edSkunk

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i think i got p = sqrt2 - 1 and A = 2-sqrt2

edit: u sure? i really dun wanna redo the q lol... gotta study phys:D
 
Last edited:
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ND

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Hahahah, i did it again, and got another answer with sqrt2, but not the same answer i got before... wtf am i doing...
 
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its not half
i got half, but like only coz im stupid
when u differentiate
u get max p
when p = root2 minus 1
and then when u sub it in
u get 0.5 something
not 0.50000
 

t-i-m-m-y

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Originally posted by worlds greatest
u get 0.5 something
not 0.50000
yeah me too.. i got close to half.. but not exactly half it was an irrational number with surds and stuff
 

walla

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area is 2 - sqrt(2)
which when you think about it is close to 0.5
you have to sub in p
 

smeyo

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what the hell are people doinf with square roots in the result, when you differentiate the equation in you get 2-2p and thus max at 1 sub that into the above formula and you get 1/2
 

smeyo

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what the hell are people doinf with square roots in the result, when you differentiate the equation in you get 2-2p and thus max at 1 sub that into the above formula and you get 1/2
 

Constip8edSkunk

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i think u did the quotient rule wrong...
from memory dA/dt =0 when p^2+2p-1 = 0 '.'p>0, p=sqrt2 - 1
 

Toodulu

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yeeea i got 1/2 as well, but i have a feeling that is wrong. i just did it and got half again. so i don't know. all the smart people seems to have a root 2 somewhere though.
 

Constip8edSkunk

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tired of physics so:

A=1-p/2+(p-1)/2(1+p)
dA/dp = -1/2 + [2(1+p)-2(p-1)]/4(1+p)^2
=1/2[ (1+p-p+1)/(1+p)^2 - (1+p)^2/(1+p)^2]
=1/2[(2-1+2p-p^2)/(1+p)^2]
=0 when
-1+2p-p^2=0
by using quadrtic formula
p={-2+/-sqrt[4+4]}/2 = -1+/-sqrt2
but p>0
so p=sqrt2-1

(show this is a maximum)

sub into original eqn

A=1-[sqrt2-1]/2+[sqrt2-2]/[2sqrt2]
=1-1/sqrt2 +1/2+1/2-1/sqrt2
=2-2sqrt2
 

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