+0

# Find the maximum value of y/x over all real numbers x and y that satisfy

+1
106
1
+461

Find the maximum value of y/x over all real numbers x and y that satisfy $$$(x - 3)^2 + (y - 3)^2 = 6.$$$

waffles  Nov 10, 2017

#1
+5931
+1

(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

hectictar  Nov 11, 2017
Sort:

#1
+5931
+1

(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

hectictar  Nov 11, 2017

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details