# What are extrema and saddle points of f(x,y)=(x+y+1)^2/(x^2+y^2+1)?

Sep 6, 2017

#### Explanation:

We have:

$f \left(x , y\right) = {\left(x + y + 1\right)}^{2} / \left({x}^{2} + {y}^{2} + 1\right)$

Step 1 - Find the Partial Derivatives

We compute the partial derivative of a function of two or more variables by differentiating wrt one variable, whilst the other variables are treated as constant. Thus:

The First Derivatives are:

${f}_{x} = \frac{\left({x}^{2} + {y}^{2} + 1\right) \left(2 \left(x + y + 1\right)\right) - \left({\left(x + y + 1\right)}^{2}\right) \left(2 x\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left({x}^{2} + {y}^{2} + 1\right) \left(x + y + 1\right) - 2 x {\left(x + y + 1\right)}^{2}}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left(x + y + 1\right) \left({x}^{2} + {y}^{2} + 1 - {x}^{2} - x y - x\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left(x + y + 1\right) \left({y}^{2} - x y - x + 1\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$

${f}_{y} = \frac{\left({x}^{2} + {y}^{2} + 1\right) \left(2 \left(x + y + 1\right)\right) - \left({\left(x + y + 1\right)}^{2}\right) \left(2 y\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left({x}^{2} + {y}^{2} + 1\right) \left(x + y + 1\right) - 2 y {\left(x + y + 1\right)}^{2}}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left(x + y + 1\right) \left({x}^{2} + {y}^{2} + 1 - {y}^{2} - x y - y\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$
$\setminus \setminus \setminus = \frac{2 \left(x + y + 1\right) \left({x}^{2} - x y - y + 1\right)}{{x}^{2} + {y}^{2} + 1} ^ 2$

The Second Derivatives (quoted) are:

${f}_{x x} = \frac{- 4 \left(- {x}^{3} - {x}^{3} y - 3 {x}^{2} y + 3 x {y}^{2} + 3 x + 3 x {y}^{3} + 3 x y + {y}^{3} + y\right)}{{x}^{2} + {y}^{2} + 1} ^ 3$

${f}_{y y} = \frac{- 4 \left(3 {x}^{3} y + {x}^{3} + 3 {x}^{2} y + 3 x y + x - x {y}^{3} - 3 x {y}^{2} + 3 y - {y}^{3}\right)}{{x}^{2} + {y}^{2} + 1} ^ 3$

The Second Partial Cross-Derivatives are:

${f}_{x y} = \frac{2 \left(- {x}^{4} - 2 {x}^{3} + 6 {x}^{2} {y}^{2} + 6 {x}^{2} y + 6 x {y}^{2} - 2 x + 1 - {y}^{4} - 2 {y}^{3} - 2 y\right)}{{x}^{2} + {y}^{2} + 1} ^ 3$

${f}_{y x} = \frac{2 \left(- {x}^{4} - 2 {x}^{3} + 6 {x}^{2} {y}^{2} + 6 {x}^{2} y + 6 x {y}^{2} - 2 x + 1 - {y}^{4} - 2 {y}^{3} - 2 y\right)}{{x}^{2} + {y}^{2} + 1} ^ 3$

Note that the second partial cross derivatives are identical due to the continuity of $f \left(x , y\right)$.

Step 2 - Identify Critical Points

A critical point occurs at a simultaneous solution of

${f}_{x} = {f}_{y} = 0 \iff \frac{\partial f}{\partial x} = \frac{\partial f}{\partial y} = 0$

i.e, when:

${f}_{x} = \frac{2 \left(x + y + 1\right) \left({y}^{2} - x y - x + 1\right)}{{x}^{2} + {y}^{2} + 1} ^ 2 = 0$
$\setminus \setminus \setminus \implies \left(x + y + 1\right) \left({y}^{2} - x y - x + 1\right) = 0$ ..... [A]

${f}_{y} = \frac{2 \left(x + y + 1\right) \left({x}^{2} - x y - y + 1\right)}{{x}^{2} + {y}^{2} + 1} ^ 2 = 0$
$\setminus \setminus \setminus \implies \left(x + y + 1\right) \left({x}^{2} - x y - y + 1\right) = 0$ ..... [B]

Solving [A] and [B] simultaneously, we gain a single solution:

$x = y = 1$

So we can conclude that there is one critical points:

$\left(1 , 1\right)$

Step 3 - Classify the critical points

In order to classify the critical points we perform a test similar to that of one variable calculus using the second partial derivatives and the Hessian Matrix.

$\Delta = H f \left(x , y\right) = | \left({f}_{x x} \setminus \setminus {f}_{x y}\right) , \left({f}_{y x} \setminus \setminus {f}_{y y}\right) | = | \left(\frac{{\partial}^{2} f}{\partial {x}^{2}} , \frac{{\partial}^{2} f}{\partial x \partial y}\right) , \left(\frac{{\partial}^{2} f}{\partial y \partial x} , \frac{{\partial}^{2} f}{\partial {y}^{2}}\right) | = {f}_{x x} {f}_{y y} - {\left({f}_{x y}\right)}^{2}$

Then depending upon the value of $\Delta$:

$\left.\begin{matrix}\Delta > 0 & \text{There is maximum if " f_(x x)<0 \\ \null & "and a minimum if " f_(x x)>0 \\ Delta<0 & "there is a saddle point" \\ Delta=0 & "Further analysis is necessary}\end{matrix}\right.$

Using custom excel macros the function values along with the partial derivative values are computed as follows: