Question

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

Answer 1

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

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

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To find the extrema, take the partial derivative with respect to `x` and `y` to see if both partial derivatives can simultaneously equal `0`.

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

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

If they simultaneously must equal `0`, they form a system of equations:

`2(2x + y + 0 = 0)`
`x + 2y + 1 = 0`

This linear system of equations, when subtracted to cancel out `y`, gives:

`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 `x` and `y` axes.

The value of `f(x,y)` at the critical point is (by plugging back into the original equation):

`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 `color(blue)(f(1/3,-2/3) = -1/3)`.

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: