Max./Min (1 Viewer)

[xwrathbringerx]

http://www.mathhelpforum.com/math-help/attachments/calculus/13605d1257045309-max-min-001.jpg

Could someone PLEASE PLEASE show me how to do the last part????

 

[addikaye03]

http://www.mathhelpforum.com/math-help/attachments/calculus/13605d1257045309-max-min-001.jpg

Could someone PLEASE PLEASE show me how to do the last part????

a) Let point vertically below O be Z which hits BA

OA=20cm (radius)


(1/2)AB=ZA (radius bisects chord at 90 degrees, <OZA=90 p degrees)<>

sin(<ZOA)=10 <ZOA="pi/6</p" therefore 20="1/2">
Since Triangle OZA congruent to triangle OBZ (SAS)


<BOZ=<ZOA=PI p congruent of angle (corresp. 6 triangle)<>

therefore <AOB=PI p 3<>
[I'm sure theres a much easier way lol]

b) Let point vertically below O be Y which hits PQ


So

Triangles POY and QOY are congruent (SAS)

Therefore PY=YQ ( corresp. sides of congruent triangles)


Therefore sin<YOQ=PY p 20<>
PY=20sin@ therefore PQ=2 (20sin@)= 40sin@ #

c) ZO-YO= Perp distance between chords

400=100+(OZ)^2 [Pythagoras]

OZ=rt(300)=10rt(3) (*)

YQ=20sin@

400=(20sin@)^2+(OY)^2

400(1-sin^2@)=(OY)^2

400cos^2@=(OY)^2

OY=20cos@

Thefore ZY=10rt(3)-20cos(@)

=10(rt3-2cos@)

I'm still going just posting this for people to try

 

[addikaye03]

d) A=h/2(a+b)

=10/2(rt3-2cos@)[40sin@+20)

=5(rt3-2cos@)20(2sin@+1)

100(1+2sin@)(rt3-2cos@)

e)

Expand above (LHS)

100(rt3-2cos@+2rt3sin@-4sin@cos@)

100rt3-200cos@+200rt3sin@-400sin(2@)

(100rt3-400sin2@)-200(cos@-rt3sin@)

cos@-rt3sin@=2sin(@-pi/6) [Using transformations of acosx+bsinx=Rcos(@-a)]

100rt3+400sin(@-pi/6)-200sin2@ (RHS)

LHS=RHS

f) =100rt3+400sin(@-pi/6)-200sin2@

d/d@=400cos(@-pi/6)-400cos2@

(Stat.Points)=0

Result follows... )

Actually i dunno if this is the right approach,

I think it might be something to do with breaking the ||gram into triangles and since A=1/2abSinC, to maximise Area SinC=90. Therefore A=(1/2)bh

That was a long Q