Posted by Knights on Friday, March 1, 2013 at 12:49pm.
Find the maximum value of y/x over all real numbers x and y that satisfy (x3)^2+(y3)^2 = 6.
It is a circle, but how do we even begin? As a matter of fact, how is there not only like 1 solution??

please help with analytic geometry if you can  Steve, Friday, March 1, 2013 at 4:16pm
it is a circle centered at (3,3) with radius √6. So, the maximum y = 3+√6.
The minimum x is likewise 3√6.
Unfortunately, the two values don't fit a point on the circle.
If you have calculus, you realize that
y/x = (3+√(6(x3)^2))/x
= (3+√(x^2+6x3))/x
y/x has a maximum where
3(√(x^2+6x3)+x1)/(x^2√(x^2+6x3)) = 0
Since the bottom is never zero, we want
√(x^2+6x3)+x1 = 0
x^2+6x3 = x^22x+1
x = 2±√2
y/x at x = 2√2 is 3+2√2 = 5.82
Not sure how you'd find this result without calculus. It's not clear that the maximum y/x occurs just near the leftmost point of the circle. It's clearly near the minimum x, but it's not obvious just how clear. 
please help with analytic geometry if you can  Reiny, Friday, March 1, 2013 at 6:18pm
Great Question!
I agree with Steve's answer, although I obtained mine using a slightly different approach
I found dy/dx of the original equation to be
dy/dx = (x3)/(3y)
I then let M = y/x
by the quotient rule,
dM/dx = (x dy/dx  y)/x^2
setting this equal to zero for a max of M
I got dy/dx = y/x (very interesting)
equating the two versions of dy/dx and simplifying gave me
x^2 + y^2 = 3x + 3y
after expanding the original circle equation and replacing x^2 + y^2
I ended up with
x+y = 4 > y = 4x
I finally subbed this back into x^2+y^2 = 3x3y
and got Steve's answer of
x = 2 ± √2
When x = 2+√2 , y = 2√2, and y/x = .5857..
which would be a minimum
when x = 2  √2 , y = 2+√2 , and y/x = 5.8284...
the maximum
Did you notice the symmetry of the x and y values for the max and min ?