answer to #5 of 03-04 Garces LT

hindi pa polished yung solution ko, so pasensya na. di pa ito papasa kay garces. haha. but since tuprio tayo... ;)

One of the acute angles of a triangle is to have a radian measure of pi/6, and the side opposite this angle is to have a length of 10 in. Prove that, of all the triangles satisfying these requirements, the one having the maximum area is isosceles.

solution:

*h/a=sin pi/6 [note that sin pi/6 = 1/2] -> h=a/2

*by cosine law,
10=a^2+b^2-2ab cos pi/6 [note that cos pi/6=sqrt(3)/2]
d/db[10=a^2+b^2-ab sqrt(3)]
0=2a da/db + 2b - sqrt(3)[b da/db + a]
0=2a da/db + 2b - b sqrt(3) da/db - a sqrt(3)
da/db=[a sqrt(3) - 2b]/[2a - b sqrt(3)]

*A=1/2 base x height [base is b, h is a/2]
d/db[A=ab/4]
dA/db=1/4(b da/db + a) [da/db is equal to the highlighted expression]
dA/db=1/4(b{[a sqrt(3) - 2b]/[2a - b sqrt(3)]} + a)

set dA/db=0 to get a critical number

[ab sqrt(3) - 2 b^2 + 2 a^2 - ab sqrt(3)]/[2a - b sqrt(3)]=0
->a^2 - b^2 = 0
(a-b)(a+b)=0
a=b <- this implies na dapat isosceles ang triangle since two sides are equal...
or a=-b (we discard this since we can't have a negative value for a length)

(insert conclusion here)