# What are the extrema and saddle points of f(x, y) = 6 sin(-x)* sin^2( y) on the interval x,y in[-pi,pi] ?

Sep 29, 2017

#### Explanation:

We have:

$f \left(x , y\right) = 6 \sin \left(- x\right) {\sin}^{2} \left(y\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 6 \sin x {\sin}^{2} y$

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:

${f}_{x} = - 6 \cos x {\sin}^{2} y$

${f}_{y} = - 6 \sin x \left(2 \sin y \cos y\right)$
$\setminus \setminus \setminus = - 6 \sin x \sin 2 y$

The Second Derivatives (quoted) are:

${f}_{x x} = 6 \sin x {\sin}^{2} y$

${f}_{y y} = - 6 \sin x \left(2 \cos 2 y\right)$
$\setminus \setminus \setminus \setminus \setminus = - 12 \sin x \cos 2 y$

The Second Partial Cross-Derivatives are:

${f}_{x y} = - 6 \cos x \sin 2 y$

${f}_{y x} = - 6 \cos x \left(2 \sin y \cos y\right)$
$\setminus \setminus \setminus \setminus = - 6 \cos x \sin 2 y$

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 = -6cosxsin^2y, = 0, ... [A]), (f_y = -6sinxsin2y, = 0, ... [B]) :}} simultaneously

Consider equation [A]

$- 6 \cos x {\sin}^{2} y = 0$

Then we have two solutions:

$\cos x = 0 \implies x = \pm \frac{\pi}{2}$
$\sin y = 0 \implies y = 0 , \pm \pi$

Now let us use Eq[B] to find the corresponding coordinate:

$x = \pm \frac{\pi}{2} \implies \sin 2 y = 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies 2 y = \pm \pi , \pm 2 \pi \implies y = \pm \frac{\pi}{2} , \pm \pi$

$y = 0 , \pm \pi \implies x \in \mathbb{R}$ (gutters)

Which gives us the following critical points:

$\left(\pm \frac{\pi}{2} , \pm \frac{\pi}{2}\right) \setminus \setminus \setminus \setminus \setminus \setminus$ (4 critical points)
$\left(\pm \frac{\pi}{2} , \pm \pi\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus$ (4 critical points)
$\left(\alpha , 0\right) \setminus \setminus \setminus \setminus \setminus \setminus \forall \alpha \in \mathbb{R} \setminus \setminus \setminus$ (gutter line)
$\left(\alpha , \pm \pi\right) \setminus \forall \alpha \in \mathbb{R} \setminus \setminus$ (2 gutter lines)

Consider equation [B]

$- 6 \sin x \sin 2 y = 0$

Then we have two solutions:

$\sin x \setminus \setminus = 0 \implies x = 0 , \pm \pi$
$\sin 2 y = 0 \implies 2 y = 0 \pm \pi , \pm 2 \pi$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies y = 0 , \pm \frac{\pi}{2} , \pm \pi$

Now let us use Eq[A] to find the corresponding coordinate@

$x = 0 , \pm \pi \implies \sin y = 0 \implies y = 0 , \pm \pi$ (repeats of above)

$y = 0 \implies x \in \mathbb{R}$ (repeat of above)

$y = \pm \frac{\pi}{2} \implies \cos x = 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \implies x = \pm \frac{\pi}{2}$ (repeats of above)

Which gives us no additional critical points:

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:

Here is a plot of the function

And the ploit with the critical points (and gutters)