This question comes from the 1991 3u HSC paper Q 7)b)
The solutions that came with this past paper book didnt really make sense for part (iii)
So can anyone else use another method to solve (iii) ?
Thanks
http://img135.imageshack.us/img135/1410/trig3ch.gif
thanks again! :)
whats the graphical method? (as u can see i dont like differentiating -.- )
Oh and whats this?
y=x^2/(x^2-9)
= (x^2-9)/(x^2-9) + 9/(x^2-9)
= 1 + 9/(x^2-9)
thanks, It was the first case ;)
one question though..
how did u get from this line:
y'' = -2(x2-16)-2 + 8x2 (x2-16)-3 = 0
to this:
Thus, -2(x2-16) + 4x2 = 0
Ok im stuck on differentiating and finding x when A'=0 ...
A = x√2500-x2
A' = ??
.....
I'm using product rule and stuck on..
x = 2500-x2
Could u guys solve it and tell me where i went wrong? thanks.
This question has got me stuck..
I've done part a) but stuck on b)
Would be great if you guys could help and show me how you did it!
Thanks!
[answer is 6.25m^2]
edit: oh crap... this is a 2u question, not 3u. Can a mod move it if its necesary?