# What are the extrema and saddle points of #f(x,y) = x^2+xy+y^2+y#?

##### 1 Answer

I found no saddle points, but there was a minimum:

#f(1/3,-2/3) = -1/3#

To find the extrema, take the partial derivative with respect to

#((delf)/(delx))_y = 2x + y#

#((delf)/(dely))_x = x + 2y + 1#

If they simultaneously must equal *system of equations*:

#2(2x + y + 0 = 0)#

#x + 2y + 1 = 0#

This *linear* system of equations, when subtracted to cancel out

#3x - 1 = 0 => color(green)(x = 1/3)#

#=> 2(1/3) + y = 0#

#=> color(green)(y = -2/3)#

Since the equations were linear, there was only one critical point, and thus only one extremum. The second derivative will tell us whether it was a maximum or minimum.

#((del^2f)/(delx^2))_y = ((del^2f)/(dely^2))_x = 2#

These second partials are in agreement, so the graph is concave up, along the *and*

The value of

#color(green)(f(1/3,-2/3)) = (1/3)^2 + (1/3)(-2/3) + (-2/3)^2 + (-2/3)#

#= 1/9 - 2/9 + 4/9 - 6/9 = color(green)(-1/3)#

Thus, we have a **minimum** of

Now, for the *cross-derivatives* to check for any saddle points that could be along a diagonal direction:

#((del^2f)/(delxdely))_(y,x) = ((del^2f)/(delydelx))_(x,y) = 1#

Since these are both in agreement as well, instead of being opposite signs, there is **no saddle point**.

We can see how this graph looks just to check: