A chord of length √3 divides a circle of unit radius into two regions. Find the rectangle of maximum area which can be inscribed in the smaller of those two regions.
If you place the origin at the center of the circle, the chord can be the line y=1/2 which is at an angle a=30
For a point (cos(a),sin(a)) on the circle pi/6<a<5pi/6, the area of the rectangle is f(a)=2cos(a)(sin(a)-1/2)
f'(a)=2cos^2(a)-2sin^2(a)+1
4sin^(a)-sin(a)-2=0
quadratic equation gives
sin(a)= (1+sqrt(33))/8 = 0.84307
a = arcsin((1+sqrt(33))/8) = 1.00297 radians
cos(a)=sqrt(1-sin^2(a)
f(a)=sqrt(15-sqrt(33))*(-3sqrt(2)+sqrt(66))/32 = 0.369001
I whipped up a desmos graph
https://www.desmos.com/calculator/giknpb21a0
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Posted by Jer
on 2024-08-22 15:02:53 |
trig + calculus sol'n
