Posted by Laquisha on Sunday, October 14, 2012 at 7:05pm.
length of rectangle -- x
width of rectangle --- y
b) radius of semicircle is y/2
c) distance = 2 lengths + 2 halfcircles
= 2x +2(1/2) π(y/2)^2
= 2x + π(y^2)/4
2x + (1/4)πy^2 = 200
times 4
8x + πy^2 = 800
x = (800 - πy^2)/8
d) -- poorly worded question.
Since they had you solve for x in c) they should have asked for the area in terms of y , not x
check your typing.
BTW, I would have defined the radius of the semicircle as r
then the width of the rectangle would be 2r, and we can avoid some of these nasty fractions.
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