Math challenging question

1.
First draw a sketch of the wall 1.8 m high at 1.2 from the building wall, for example, see:
http://i263.photobucket.com/albums/ii157/mathmate/math/ladder.jpg
where t=angle of elevation of ladder, L=total length.
By simple trigonometry for the height where the ladder touches the building, we can deduce that
Lsin(t)=1.8+1.2tan(t), or
L=1.8/sin(t)+1.2/cos(t)
differentiate w.r.t. t and equate to zero for a minimum:
we get
dL/dt=1.2tan(t)/cos(t)-1.8cot(t)/sin(t)=0
and note that the function is singular at t=0 and t=pi/2.
Multiply by cot(t) to get
1.2tan(t)-1.8cot(t)^2
tan(t)=(1.8/1.2)^(1/3)
tan(t)=1.1447142...
t=0.852770877...
See the plots of dL/dt w.r.t. t
http://i263.photobucket.com/albums/ii157/mathmate/math/ladder1.png
http://i263.photobucket.com/albums/ii157/mathmate/math/ladder2.jpg
Substitute in the L(t) expression above to get
L(0.852...)
L=1.8/sin(0.852770877)+1.2/cos(0.852770877)
=4.214 m.

2008-05-19 05:19:24 補充：
2.
y=asin(x)+bcos(x)
dy/dx=acos(x)-bsin(x)=0 (at max. or min.)
from which tan(x)=a/b, sin(x)=a/sqrt(a^2+b^2), cos(x)=b/sqrt(a^2+b^2)
y
=asin(x)+bcos(x)
=(a^2+b^2)/sqrt(a^2+b^2)
=+/- sqrt(a^2+b^2)