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??
Both Steve and I have answered your question
http://www.jiskha.com/display.cgi?id=1362160145
What part of the answers did you not like ?
I think this is the third time you have posted the same question, please check your previous posts before re-posting.
Oh I'm sorry, I thought no had done so, so thanks.
To find the maximum value of y/x, we need to understand the equation (x-3)^2 + (y-3)^2 = 6, which represents a circle. This equation describes all the points (x, y) that are a distance of 6 from the center of the circle at point (3, 3).
To start solving this problem, let's rewrite the equation in terms of y, and then we can differentiate it implicitly to find the extreme values.
(x-3)^2 + (y-3)^2 = 6, which can be rewritten as:
(y-3)^2 = 6 - (x-3)^2.
Now, when finding the maximum value of y/x, we can assume x ≠ 0, as division by zero is not possible. To proceed, we can divide both sides of the equation by x^2.
((y-3)^2) / (x^2) = (6 - (x-3)^2) / (x^2).
Now, taking the derivative of both sides with respect to x implicitly, we get:
[2(y-3)(y')x - 2(x-3)] / (x^3) = [(6 - (x-3)^2)' / (x^2)].
Simplifying further:
2(y-3)(y')x - 2(x-3) = (6x^2 - 12x + 27 - x^2 + 6x - 9) / (x^3),
2(y-3)(y')x - 2(x-3) = (5x^2 - 12x + 18) / (x^3).
Now, to find the maximum value of y/x, we can solve for y' (the derivative of y with respect to x) and set it equal to zero.
2(y-3)(y')x - 2(x-3) = 0,
(y-3)(y')x - (x-3) = 0,
(y-3)(y')x = x-3,
y' = (x-3) / ((y-3)x).
To find the extreme values, we need to evaluate y'/x at the points (x, y) where the equation (x-3)^2 + (y-3)^2 = 6 is satisfied.
Now, substituting y = mx into (x-3)^2 + (y-3)^2 = 6, we get:
(x-3)^2 + (mx-3)^2 = 6,
x^2 - 6x + 9 + m^2x^2 - 6mx + 9 = 6,
(1+m^2)x^2 - 6(m+1)x + 12 = 0.
To have real solutions for x, the discriminant of this quadratic equation should be greater than or equal to zero.
The discriminant, D = (-6(m+1))^2 - 4(1+m^2)(12),
D = 36(m+1)^2 - 48(1+m^2),
D = 36(m^2 + 2m + 1) - 48 - 48m^2,
D = -12m^2 + 72m -12.
For the discriminant to be greater than or equal to zero:
-12m^2 + 72m -12 ≥ 0,
12m^2 - 72m + 12 ≤ 0,
m^2 - 6m + 1 ≤ 0.
Now, to solve the inequality, we factor the quadratic expression:
(m-3+√5)(m-3-√5) ≤ 0.
Using the principle that the product of two factors is less than or equal to zero if and only if one factor is positive and the other is negative, we have two cases:
Case 1: (m-3+√5) ≥ 0 and (m-3-√5) ≤ 0,
m-3+√5 ≥ 0, implies m ≥ 3 - √5,
m-3-√5 ≤ 0, implies m ≤ 3 + √5.
Case 2: (m-3+√5) ≤ 0 and (m-3-√5) ≥ 0,
m-3+√5 ≤ 0, implies m ≤ 3 - √5,
m-3-√5 ≥ 0, implies m ≥ 3 + √5.
Therefore, the solution to m lies in the range:
3 + √5 ≤ m ≤ 3 - √5.
Now that we have the range for m, we can find the maximum value of y/x by substituting these endpoints into y' = (x-3) / ((y-3)x).
Let's evaluate y'/x at these endpoints:
For m = 3 + √5:
y' = (x-3) / ((y-3)x) = (x-3) / ((3 + √5 - 3)x) = (x-3) / ((√5)x) = 1/√5.
For m = 3 - √5:
y' = (x-3) / ((y-3)x) = (x-3) / ((3 - √5 - 3)x) = (x-3) / ((-√5)x).
As we are interested in the maximum value of y/x, we need to find the maximum value of y', which is 1/√5, corresponding to m = 3 + √5.
Therefore, the maximum value of y/x over all real numbers x and y that satisfy the equation (x-3)^2 + (y-3)^2 = 6 is 1/√5.
Note: Although there is only one maximum value of y/x, the equation represents a circle, and there are infinitely many points (x, y) that satisfy the equation. Each point represents a unique value of y/x.