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

[VanCarBus]

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

 

[symo]

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

 

[VanCarBus]

same, @ = tan-1 (1/p)

 

[Constip8edSkunk]

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.

 

[Toodulu]

was the last part 1/2?

 

[ND]

I got some answer with sqrt2's and stuff.

 

[felix_js]

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

 

[jogloran]

I got 1/2, and only got a = arctan(1/p)...

 

[ND]

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

 

[Constip8edSkunk]

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

 

[ND]

Hahahah, i did it again, and got another answer with sqrt2, but not the same answer i got before... wtf am i doing...

 

[sugaryblue]

Originally posted by VanCarBus
Huh?
I got a full on blackout on this question. Anyone else?

Couldn't do the third part. i think that was it

question 7 was bloody hard!

 

[worlds greatest]

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]

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]

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

 

[smeyo]

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]

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]

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]

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]

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