"Find the maximum value of y/x over all real numbers x and y that satisfy (x - 3)^2 + (y - 3)^2 = 6."
I tried simplifying it to xy(x+y-12), went nowhere.
9 answers
I also know that when drawn on a coordinate plane the figure is a circle.
where does the tangent to the circle from the origin hit the circle?
draw circle, center at (3,3) radius sqrt 6
draw line (hypotenuse it turns out) from (0,0) to (3,3), length is 3 sqrt 2
draw our tangent, hits circle at 90 degree angle to radius to that point, length is h
so
h^2 = 18 + 6 = 24
h = 2 sqrt 6
I think you can figure out the geometry from there to find the tangent from the origin which is max y/x
draw line (hypotenuse it turns out) from (0,0) to (3,3), length is 3 sqrt 2
draw our tangent, hits circle at 90 degree angle to radius to that point, length is h
so
h^2 = 18 + 6 = 24
h = 2 sqrt 6
I think you can figure out the geometry from there to find the tangent from the origin which is max y/x
whoops, sorry
h^2 = 18 - 6 = 12
h = 2 sqrt 3
h^2 = 18 - 6 = 12
h = 2 sqrt 3
angle from x axis to tangent = 45 + sin^-1 ( sqrt 6/3sqrt2)
= 45 + 35.3 = 80.26 deg
x = 2 sqrt 3 * cos 80.26
y = 2 sqrt 3 * sin 80.26
y/x = tan 80.26 = 5.82
= 45 + 35.3 = 80.26 deg
x = 2 sqrt 3 * cos 80.26
y = 2 sqrt 3 * sin 80.26
y/x = tan 80.26 = 5.82
The line y=kx is tangent to the circle in two places. This will occur in two places, one with maximum slope, and one with minimum slope (y/x).
As a function of x, y = 3±√(6-(x-3)^2)
Clearly the maximum slope is achieved on the upper half-circle.
kx = -(x-3)/(y-3)
= -(x-3)/√(6-(x-3)^2)
k = -(x-3) / x√(6-(x-3)^2)
Now just find where k is a maximum.
y/x = (3+√(6-(x-3)^2))/x
So, find the max of that.
I get the point (0.587,3.421), where k = 5.828
http://www.wolframalpha.com/input/?i=plot+%28x-3%29^2+%2B+%28y-3%29^2+%3D+6%2C+y%3D5.828x
As a function of x, y = 3±√(6-(x-3)^2)
Clearly the maximum slope is achieved on the upper half-circle.
kx = -(x-3)/(y-3)
= -(x-3)/√(6-(x-3)^2)
k = -(x-3) / x√(6-(x-3)^2)
Now just find where k is a maximum.
y/x = (3+√(6-(x-3)^2))/x
So, find the max of that.
I get the point (0.587,3.421), where k = 5.828
http://www.wolframalpha.com/input/?i=plot+%28x-3%29^2+%2B+%28y-3%29^2+%3D+6%2C+y%3D5.828x
Nice one, Damon. My calculations get kind of complicated. Your is much, much simpler.
At least we agree on the maximum y/x = 5.82
At least we agree on the maximum y/x = 5.82
Could have saved myself some work if I'd kept that in mind!
I started with Calculus first as well, but changed my mind after drawing the picture.