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??
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.
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 ?
To find the maximum value of y/x, we need to first understand the geometric interpretation of the given equation, (x-3)^2 + (y-3)^2 = 6.
This equation represents a circle centered at the point (3, 3) with a radius of √6. The equation tells us that the sum of the squares of the distances between any point (x, y) on the circle and the point (3, 3) is equal to 6.
Now, let's consider the condition y/x. We want to maximize this ratio. However, we need to take into account that the denominator, x, must be nonzero since division by zero is undefined.
To find the maximum value, we can use a little bit of algebraic manipulation. Rearrange the equation (x-3)^2 + (y-3)^2 = 6 to solve for y:
(y-3)^2 = 6 - (x-3)^2
Taking the square root of both sides:
y - 3 = ±√(6 - (x-3)^2)
Now, solving for y:
y = 3 ± √(6 - (x-3)^2)
Since we're interested in maximizing y/x, we substitute the expression for y in terms of x:
y/x = (3 ± √(6 - (x-3)^2))/x
Now, we need to analyze the behavior of y/x as x varies. Let's consider two cases:
1. x > 0:
If x > 0, there are two possible situations:
a. (3 ± √(6 - (x-3)^2))/x: The numerator is positive (since both 3 and √(6 - (x-3)^2) are positive), and dividing a positive number by a positive number gives a positive result.
b. (3 ± √(6 - (x-3)^2))/x: The numerator is negative (since both 3 and √(6 - (x-3)^2) are positive), and dividing a negative number by a positive number gives a negative result.
2. x < 0:
If x < 0, there are also two possible situations:
a. (3 ± √(6 - (x-3)^2))/x: The numerator is positive (since both 3 and √(6 - (x-3)^2) are positive), and dividing a positive number by a negative number gives a negative result.
b. (3 ± √(6 - (x-3)^2))/x: The numerator is negative (since both 3 and √(6 - (x-3)^2) are positive), and dividing a negative number by a negative number gives a positive result.
Now, let's plot the graph of the circle (x-3)^2 + (y-3)^2 = 6 and analyze the behavior of y/x for each possible situation: