# 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]# ?

We have:

# f(x,y) = 6sin(-x)sin^2(y) #

# \ \ \ \ \ \ \ \ \ \ \ = -6sinxsin^2y #

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 = -6cosxsin^2y#

# f_y = -6sinx(2sinycosy) #

# \ \ \ = -6sinxsin2y #

The Second Derivatives (quoted) are:

# f_(x x) = 6sinxsin^2y #

# f_(yy) = -6sinx(2cos2y) #

# \ \ \ \ \ = -12sinxcos2y #

The Second Partial Cross-Derivatives are:

# f_(xy) = -6cosxsin2y #

# f_(yx) = -6cosx(2sinycosy) #

# \ \ \ \ = -6cosxsin2y #

Note that the second partial cross derivatives are identical due to the continuity of

Step 2 - Identify Critical Points

A critical point occurs at a simultaneous solution of

# f_x = f_y = 0 iff (partial f) / (partial x) = (partial f) / (partial y) = 0#

i.e, when:

# {: (f_x = -6cosxsin^2y, = 0, ... [A]), (f_y = -6sinxsin2y, = 0, ... [B]) :}}# simultaneouslyConsider equation [A]

# -6cosxsin^2y = 0 # Then we have two solutions:

# cosx = 0 => x = +- pi/2 #

# sin y = 0 => y = 0, +- pi# Now let us use Eq[B] to find the corresponding coordinate:

# x = +-pi/2 => sin2y = 0 #

# \ \ \ \ \ \ \ \ => 2y = +-pi, +- 2pi => y = +- pi/2, +-pi #

# y=0,+-pi => x in RR # (gutters)Which gives us the following critical points:

# (+-pi/2, +-pi/2) \ \ \ \ \ \ # (4 critical points)

# (+-pi/2, +-pi) \ \ \ \ \ \ \ \ # (4 critical points)

# (alpha, 0) \ \ \ \ \ \AA alpha in RR \ \ \ # (gutter line)

# (alpha, +-pi) \ AA alpha in RR \ \ # (2 gutter lines)Consider equation [B]

# -6sinxsin2y = 0 # Then we have two solutions:

# sinx \ \= 0 => x = 0,+- pi #

# sin2y = 0 => 2y = 0+- pi, +-2pi#

# \ \ \ \ \ \ \ \ => y = 0, +-pi/2, +- pi # Now let us use Eq[A] to find the corresponding coordinate@

# x=0,+-pi => siny=0 => y=0,+-pi # (repeats of above)

# y=0 => x in RR # (repeat of above)

# y = +-pi/2 => cosx = 0 #

# \ \ \ \ \ \ \ \ => x=+-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(x,y) = | ( f_(x x) \ \ f_(xy) ) , (f_(yx) \ \ f_(yy)) | = | ((partial^2 f) / (partial x^2),(partial^2 f) / (partial x partial y)), ((partial^2 f) / (partial y partial x), (partial^2 f) / (partial y^2)) | = f_(x x)f_(yy)-(f_(xy))^2 # Then depending upon the value of

#Delta# :

# {: (Delta>0, "There is maximum if " f_(x x)<0),(, "and a minimum if " f_(x x)>0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :} # 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)

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To find the extrema and saddle points of ( f(x, y) = 6 \sin(-x) \cdot \sin^2(y) ) on the interval ( [-\pi, \pi] ), we first need to find the critical points by taking the partial derivatives of ( f ) with respect to ( x ) and ( y ) and setting them equal to zero. Then, we classify these points using the second partial derivative test.

Taking the partial derivatives:

[ \frac{\partial f}{\partial x} = -6 \cos(-x) \cdot \sin^2(y) ] [ \frac{\partial f}{\partial y} = 12 \sin(-x) \cdot \sin(y) \cdot \cos(y) ]

Setting them equal to zero:

[ -6 \cos(-x) \cdot \sin^2(y) = 0 \quad \text{and} \quad 12 \sin(-x) \cdot \sin(y) \cdot \cos(y) = 0 ]

Solving these equations for ( x ) and ( y ) on the interval ( [-\pi, \pi] ), we find the critical points.

For the second derivative test, we compute the second partial derivatives:

[ \frac{\partial^2 f}{\partial x^2} = 6 \sin(-x) \cdot \sin^2(y) ] [ \frac{\partial^2 f}{\partial y^2} = 12 \sin(-x) \cdot \sin(y) \cdot \cos(y) ] [ \frac{\partial^2 f}{\partial x \partial y} = -24 \cos(-x) \cdot \sin(y) \cdot \cos(y) ]

Then we evaluate these at the critical points and determine the nature of the critical points based on the signs of these second partial derivatives.

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