# What are the values and types of the critical points, if any, of f(x, y) = xy^2+xy-3x^2y-7x?

Apr 28, 2018

Saddle points at $\left(0 , - \frac{1}{2} - \frac{\sqrt{29}}{2}\right)$ and $\left(0 , - \frac{1}{2} + \frac{\sqrt{29}}{2}\right)$

#### Explanation:

We have a 3D surface defined by:

$f \left(x , y\right) = x {y}^{2} + x y - 3 {x}^{2} y - 7 x$

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:

${f}_{x} \setminus = \frac{\partial f}{\partial x} \setminus \setminus = {y}^{2} + y - 6 x y - 7$
${f}_{y} \setminus = \frac{\partial f}{\partial y} \setminus \setminus = 2 x y + x - 3 {x}^{2}$

The Second Derivatives are:

${f}_{x x} = \frac{{\partial}^{2} f}{\partial {x}^{2}} = - 6 y$
${f}_{y y} = \frac{{\partial}^{2} f}{\partial {y}^{2}} = 2 x$

The Second Partial Cross-Derivatives are:

${f}_{x y} = \frac{{\partial}^{2} f}{\partial x \partial y} = 2 y + 1 - 6 x$
${f}_{y x} = \frac{{\partial}^{2} f}{\partial y \partial x} = 2 y + 1 - 6 x$

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:

${y}^{2} + y - 6 x y - 7 = 0$ ..... [A]
$2 x y + x - 3 {x}^{2} = 0$ ..... [B]

From [B], we have:

$x \left(2 y + 1 - 3 x\right) = 0 \implies x = 0 , \frac{1}{3} \left(2 y + 1\right)$

From [A], we have:

$x = 0 \implies {y}^{2} + y - 7 = 0$
$\therefore y = - \frac{1}{2} \pm \frac{\sqrt{29}}{2}$

$x = \frac{1}{3} \left(2 y + 1\right) \implies {y}^{2} + y - 6 \left(\frac{1}{3} \left(2 y + 1\right)\right) y - 7 = 0$
$\therefore {y}^{2} + y - 4 {y}^{2} - 2 y - 7 = 0$
$\therefore 3 {y}^{2} + y + 7 = 0$ having no real solutions

So we can conclude that there are two critical points:

$\left(0 , - \frac{1}{2} - \frac{\sqrt{29}}{2}\right)$ and $\left(0 , - \frac{1}{2} + \frac{\sqrt{29}}{2}\right)$

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:

And we can confirm these results graphically