# Identify the Critical Points for the function? :  f(x,y) = 3xy-x^3-3y^2

Aug 3, 2017

The saddle point is $\left(0 , 0\right)$ and the local maximum is $\left(\frac{1}{2} , \frac{1}{4}\right)$

#### Explanation:

Our function is

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

The partial derivatives are

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

${f}_{y} \left(x , y\right) = 3 x - 6 y$

${f}_{x x} \left(x , y\right) = - 6 x$

${f}_{y y} \left(x , y\right) = - 6$

${f}_{x y} \left(x , y\right) = 3$

${f}_{y x} \left(x , y\right) = 3$

We look for the critical points

${f}_{x} \left(x , y\right) = 3 y - 3 {x}^{2} = 0$, $\implies$, $y = {x}^{2}$...........$\left(1\right)$

${f}_{y} \left(x , y\right) = 3 x - 6 y = 0$, $\implies$, $y = \frac{x}{2}$.................$\left(2\right)$

We solve for $x$ and $y$ in equations $\left(1\right)$ and $\left(2\right)$

${x}^{2} = \frac{x}{2}$

${x}^{2} - \frac{x}{2} = 0$

$x \left(x - \frac{1}{2}\right) = 0$

$x = 0$ and $x = \frac{1}{2}$

Therefore, the critical points are

$\left(0 , 0\right)$ and $\left(\frac{1}{2} , \frac{1}{4}\right)$

Therefore,

${f}_{x x} \left(0 , 0\right) = - 6 \cdot 0 = 0$ and ${f}_{x x} \left(\frac{1}{2} , \frac{1}{4}\right) = - 6 \cdot \frac{1}{2} = - 3$

${f}_{y y} \left(0 , 0\right) = - 6$ and ${f}_{y y} \left(\frac{1}{2} , \frac{1}{4}\right) = - 6$

${f}_{x y} \left(0 , 0\right) = 3$ and ${f}_{x y} \left(\frac{1}{2} , \frac{1}{4}\right) = 3$

To perform the second derivative test, we calculate the determinant,

$D \left(0 , 0\right) = | \left(0 , 3\right) , \left(3 , - 6\right) |$ $= \left(0\right) \cdot \left(- 6\right) - \left(3\right) \cdot \left(3\right) = - 9 < 0$, this represents a saddle point

$D \left(\frac{1}{2} , \frac{1}{4}\right) = | \left(- 3 , 3\right) , \left(3 , - 6\right) |$ $= \left(- 3\right) \cdot \left(- 6\right) - \left(3\right) \cdot \left(3\right) = 18 - 9 = 9 > 0$, this represents a local maximum as ${f}_{x x} \left(\frac{1}{2} , \frac{1}{4}\right) < 0$

Aug 3, 2017

$\left(0 , 0\right)$- Saddle Point

$\left(\frac{1}{2} , \frac{1}{4}\right)$ - Maximum

#### Explanation:

We have:

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

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:

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

The Second Derivatives are:

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

The Second Partial Cross-Derivatives are:

${f}_{x y} = \frac{{\partial}^{2} f}{\partial x \partial y} = 3$
${f}_{y x} = \frac{{\partial}^{2} f}{\partial y \partial x} = 3$

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:

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

From [A] we have: $y = {x}^{2}$ so substituting for $y$ in [B] we get:

$3 x - 6 {x}^{2} = 0 \implies 2 {x}^{2} - x = 0$
$\therefore x \left(2 x - 1\right) = 0$
$\therefore x = 0 , \frac{1}{2}$

And the using $y = {x}^{2}$ we have:

$x = 0 \setminus \setminus \implies y = 0$
$x = \frac{1}{2} \implies y = \frac{1}{4}$

So we can conclude that there are two critical points:

$\left(0 , 0\right)$; and $\left(\frac{1}{2} , \frac{1}{4}\right)$

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: