please help with analytic geometry if you can

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Find the maximum value of y/x over all real numbers x and y that satisfy (x-3)^2+(y-3)^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 -

    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-(x-3)^2))/x
    = (3+√(-x^2+6x-3))/x
    y/x has a maximum where
    3(√(-x^2+6x-3)+x-1)/(x^2√(-x^2+6x-3)) = 0

    Since the bottom is never zero, we want

    √(-x^2+6x-3)+x-1 = 0
    -x^2+6x-3 = x^2-2x+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 -

    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 = (x-3)/(3-y)

    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 = 4-x
    I finally subbed this back into x^2+y^2 = 3x-3y
    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 ?

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