Question

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??

Answer 1

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.

Answer 2

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 ?