x^2 + y^2 − 30x − 40y + 242 = 0
(x-15)^2 + (y-20)^2 = 383
This is a circle centered at (15,20) with radius √383
If x=0 then y/x is maximized.
Note that the circle intersects the y-axis (x=0) in two places.
y^2 − 40y + 242 = 0 after substituting 0 for x.
y = 20 ± √158
So maximum y/x is when (x,y) = (0, 20+√158) ~ (0, 32.5698)
or maximum y/x is when (x,y) = (0, 20-√158) ~ (0, 7.43019)
The requested ratio of the maximum of y/x is infinity.
(At least in the limit as x approaches 0)
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Posted by Larry
on 2024-11-01 12:18:31 |
solution