What is the maximum value of the function 5xy+2x+2y subject to (x^2)(y^2)=36?
+130511 Assuming xy > 0, we have
(x^2)(y^2)=36 → [ xy = 6 ] → [ y = 6/x ]
So
5xy+2x+2y =
30 + 2x + 12/x
And taking the derivative and setting it to 0, we have
2 - 12/x2 = 0
1 - 6/x2 = 0 rearranging, we have
x2 = 6
x = ±√6
And taking the second derivative, we have
12x-3
Putting in √6 produces a positive indicating that the graph is concave up at this point. And putting in -√6 produces a negative which indicates the graph is concave down at this point.
If we restrict ourselves to x < 0 then we have a relative maximum at x = -√6 y = 20.20
And if we restirct ourselves to x > 0 we a relative minimum at √6 y = 39.8
The graph actually has no absolute max
Here's the graph......https://www.desmos.com/calculator/posrwbc4ax
+130511 Assuming xy > 0, we have
(x^2)(y^2)=36 → [ xy = 6 ] → [ y = 6/x ]
So
5xy+2x+2y =
30 + 2x + 12/x
And taking the derivative and setting it to 0, we have
2 - 12/x2 = 0
1 - 6/x2 = 0 rearranging, we have
x2 = 6
x = ±√6
And taking the second derivative, we have
12x-3
Putting in √6 produces a positive indicating that the graph is concave up at this point. And putting in -√6 produces a negative which indicates the graph is concave down at this point.
If we restrict ourselves to x < 0 then we have a relative maximum at x = -√6 y = 20.20
And if we restirct ourselves to x > 0 we a relative minimum at √6 y = 39.8
The graph actually has no absolute max
Here's the graph......https://www.desmos.com/calculator/posrwbc4ax
