What is the ratio of the area of S2β to the area of S1β?
Answer Choices:
A. 98
B. 99
C. 100
D. 101
E. 102 Solution:
For j=1 and 2 , the given inequality is equivalent to
j+x2+y2β€10j(x+y)
or to
(xβ210jβ)2+(yβ210jβ)2β€2102jββj
provided that x+y>0. These inequalities define regions bounded by circles. For j=1 the circle has center (5,5) and radius 7 . For j=2 the circle has center (50,50) and radius 4998β. In each case the center is on the line y=x in the first quadrant, and the radius is less than the distance from the center to the origin. Thus x+y>0 at each interior point of each circle, as was required to ensure the equivalence of the inequalities. The squares of the radii of the circles are 49 and 4998 for j=1 and 2 , respectively. Therefore the ratio of the area of S2β to that of S1β is (4998Ο)/(49Ο)=102β.