# Thread: Maximum/minimum values

1. ## Maximum/minimum values

Find the local max/min values and saddle points:

$f(x,y) = (1+xy)(x+y)$

taking the partial derivatives I get

$f_x = y(x+y) + (1+xy) \ \ \mbox{and} \ \ f_y = x(x+y)+(1+xy)$

$f_x = y(x+y) + (1+xy) = 0$

$y(x+y) = - (1+xy)$

$xy+y^2 = -1-xy$

$x = -\frac{y}{2} - \frac{1}{2y}$

taking:

$f_y = x(x+y)+(1+xy) = 0$

$y = -\frac{x}{2} - \frac{1}{2x}$

if I make the replacement I get:

$y= \left( \frac{y}{4} - \frac{1}{4y} - \left( \frac{1}{-y-\frac{1}y} \right) \right) = \frac{5y}{4} + \frac{3}{4y}$

is this correct?

2. Originally Posted by lllll
Find the local max/min values and saddle points:

$f(x,y) = (1+xy)(x+y)$

taking the partial derivatives I get

$f_x = y(x+y) + (1+xy) \ \ \mbox{and} \ \ f_y = x(x+y)+(1+xy)$

$f_x = y(x+y) + (1+xy) = 0$

$y(x+y) = - (1+xy)$

$xy+y^2 = -1-xy$

$x = -\frac{y}{2} - \frac{1}{2y}$

taking:

$f_y = x(x+y)+(1+xy) = 0$

$y = -\frac{x}{2} - \frac{1}{2x}$

if I make the replacement I get:

$y= \left( \frac{y}{4} - \frac{1}{4y} - \left( \frac{1}{-y-\frac{1}y} \right) \right) = \frac{5y}{4} + \frac{3}{4y}$

is this correct?
You're hoeing the hard road.

$\frac{\partial f}{\partial x} = y^2 + 2xy + 1 = 0$ .... (1)

$\frac{\partial f}{\partial y} = x^2 + 2xy + 1 = 0$ .... (2)

(1) - (2): $y^2 - x^2 = 0 \Rightarrow (y - x)(y + x) = 0 \Rightarrow y = \pm x$.

Case 1: y = x

Substitute into (1): $3x^2 + 1 = 0$. No real solution.

Case 2: y = -x

Substitute into (1): $-x^2 + 1 = 0 \Rightarrow x = \pm 1$.

Extremum solutions are (1, -1) and (-1, 1).

Now you have to test their nature.