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??
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To find the maximum value of y/x, we need to determine the maximum slope of the line passing through the origin (0,0) and any point (x, y) on the given circle equation, (x-3)^2+(y-3)^2 = 6.
To begin, let's rewrite the equation in the general form:
x^2 - 6x + y^2 - 6y + 9 - 6 = 0
x^2 - 6x + y^2 - 6y + 3 = 0
Notice that this equation represents a circle centered at (3,3) with a radius of sqrt(6). Since we want to find the maximum value of y/x, we can use the concept of tangents to the circle.
First, let's differentiate the equation with respect to x to find the slope of the tangent line, which will be the maximum value of y/x:
2x - 6 + 2y * (dy/dx) - 6(dy/dx) = 0
(2y - 6)(dy/dx) = 6 - 2x
dy/dx = (6 - 2x) / (6 - 2y)
To find the maximum value of y/x, we can set the derivative dy/dx equal to zero:
(6 - 2x) / (6 - 2y) = 0
Solving for y, we have:
6 - 2x = 0
2x = 6
x = 3
This implies that y can take any value since the value of x does not affect y. Therefore, there isn't a unique solution for y/x. The value of y/x will be maximum when y is large and positive or small and negative; likewise, it will be minimum when y is large and negative or small and positive.
In conclusion, there are infinitely many points on the circle where y/x can take its maximum value.