Question

"Find the maximum value of y/x over all real numbers x and y that satisfy (x - 3)^2 + (y - 3)^2 = 6."
I tried simplifying it to xy(x+y-12), went nowhere.

Answer 1

I also know that when drawn on a coordinate plane the figure is a circle.

Answer 2

where does the tangent to the circle from the origin hit the circle?

Answer 3

draw circle, center at (3,3) radius sqrt 6

draw line (hypotenuse it turns out) from (0,0) to (3,3), length is 3 sqrt 2

draw our tangent, hits circle at 90 degree angle to radius to that point, length is h

so
h^2 = 18 + 6 = 24
h = 2 sqrt 6

I think you can figure out the geometry from there to find the tangent from the origin which is max y/x

Answer 4

whoops, sorry
h^2 = 18 - 6 = 12
h = 2 sqrt 3

Answer 5

angle from x axis to tangent = 45 + sin^-1 ( sqrt 6/3sqrt2)

= 45 + 35.3 = 80.26 deg

x = 2 sqrt 3 * cos 80.26
y = 2 sqrt 3 * sin 80.26

y/x = tan 80.26 = 5.82

Answer 6

The line y=kx is tangent to the circle in two places. This will occur in two places, one with maximum slope, and one with minimum slope (y/x).

As a function of x, y = 3±√(6-(x-3)^2)
Clearly the maximum slope is achieved on the upper half-circle.

kx = -(x-3)/(y-3)
= -(x-3)/√(6-(x-3)^2)

k = -(x-3) / x√(6-(x-3)^2)

Now just find where k is a maximum.

y/x = (3+√(6-(x-3)^2))/x
So, find the max of that.
I get the point (0.587,3.421), where k = 5.828

http://www.wolframalpha.com/input/?i=plot+%28x-3%29^2+%2B+%28y-3%29^2+%3D+6%2C+y%3D5.828x

Answer 7

Nice one, Damon. My calculations get kind of complicated. Your is much, much simpler.

At least we agree on the maximum y/x = 5.82

Answer 8

Could have saved myself some work if I'd kept that in mind!

Answer 9

I started with Calculus first as well, but changed my mind after drawing the picture.