+647 Find the maximum value of y/x over all real numbers x and y that satisfy \(\[(x - 3)^2 + (y - 3)^2 = 6.\]\)
+9479 (x + 3)2 + (y - 3)2 = 6 Let's solve this for y .
(y - 3)2 = 6 - (x + 3)2
y - 3 = ±√[ 6 - (x + 3)2 ]
y = ±√[ 6 - (x + 3)2 ] + 3 So....using this value for y....
\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)
We can say
\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\) and we want to know the maximum Y value here.
We can get an approximation by looking at a graph. (about 5.828)
If you take the derivative and set it = 0, you will get x = √2 - 2
Then plug this in for x and we can find that the exact maximum value = 3 + 2√2
+9479 (x + 3)2 + (y - 3)2 = 6 Let's solve this for y .
(y - 3)2 = 6 - (x + 3)2
y - 3 = ±√[ 6 - (x + 3)2 ]
y = ±√[ 6 - (x + 3)2 ] + 3 So....using this value for y....
\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)
We can say
\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\) and we want to know the maximum Y value here.
We can get an approximation by looking at a graph. (about 5.828)
If you take the derivative and set it = 0, you will get x = √2 - 2
Then plug this in for x and we can find that the exact maximum value = 3 + 2√2
