To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2(x+4)+6xy=2x+8+6xy$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=-4(y-1)+3x^2=3x^2-4y+4$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2+6y$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-4$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=6x$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=2x+8+6xy=0\Rightarrow 2x+6xy=-8$

$\displaystyle 2x+6xy=-8\Rightarrow x+3xy=-4\Rightarrow x(1+3y)=-4$

$\displaystyle x=\frac{-4}{1+3y}$

$\displaystyle f_y(x,y)=3x^2-4y+4=0\Rightarrow 3\left ( \frac{-4}{1+3y} \right )^2-4y+4=0$

$\displaystyle 3\left ( \frac{16}{(1+3y)^2} \right )-4y+4=0$

$\displaystyle \frac{48}{(1+3y)^2}=4y-4$

$\displaystyle 48=(4y-4)(1+3y)^2$

$\displaystyle (4y-4)(1+3y)^2-48=0$

$\displaystyle (4y-4)(1+6y+9y^2)-48=0$

$\displaystyle 4y+24y^2+36y^3-4-24y-36y^2-48=0$

$\displaystyle 36y^3-12y^2-20y-52=0$

There is only one real value of $\displaystyle y$; $\displaystyle y\approx1.429$

We find the corresponding value of $\displaystyle x$ using $\displaystyle x=\frac{-4}{1+3y}$ (found by rearranging the first derivative)

$\displaystyle x(1.429)=\frac{-4}{1+3(1.429)}\approx -0.757$

There is a critical point at $\displaystyle (1.429,-0.757)$.  We need to determine if the critical point is a maximum or minimum using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(2+6y)(-4)-(6x)^2$

$\displaystyle D(x,y)=-8-24y-36x^2$

At $\displaystyle (1.429,-0.757)$,

$\displaystyle D(x,y)=-8-24(-0.757)-36(1.429)^2=-63.35< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (1.429,-0.757)$ is a saddle point.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=6x^2-4xy-7y$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=4y^3-2x^2-7x$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x-4y$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=-4x-7$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=6x^2-4xy-7y=0\Rightarrow 6x^2-y(4x-7)=0$

$\displaystyle 6x^2=y(4x-7)\Rightarrow y=\frac{6x^2}{4x-7}$

$\displaystyle f_y(x,y)=4y^3-2x^2-7x=0\Rightarrow 4\left ( \frac{6x^2}{4x-7} \right )^3-2x^2-7x=0$

$\displaystyle 4(6x^2)^3-2x^2(4x-7)^3-7x(4x-7)^3=0(4x-7)^3$

$\displaystyle 4(6x^2)^3-(2x^2+7x)(4x-7)^3=0$

$\displaystyle 864x^6-128x^5+224x^4+1176x^3-3430x^2+2401x=0$

The real values of $\displaystyle x$ are $\displaystyle x=0$ and $\displaystyle x=-1.6$

We find the corresponding value of $\displaystyle y$ using $\displaystyle y=\frac{6x^2}{4x-7}$ (found by rearranging the first derivative)

$\displaystyle y(0)=\frac{6(0)^2}{4(0)-7}=0$

$\displaystyle y(-1.6)=\frac{6(-1.6)^2}{4(-1.6)-7}=\frac{6*2.56}{-1.6-7}\approx -1.786$

There are critical points at $\displaystyle (0,0)$ and $\displaystyle (-1.6,-1.786)$.  We need to determine if the critical points are maxima or minima using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(12x-4y)(12y^2)-(-4x-7)^2$

 At $\displaystyle (0,0)$,

$\displaystyle D(x,y)=(12(0)-4(0))(12(0)^2)-(-4(0)-7)^2=0-(-7)^2=-49< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (0,0)$ is a saddle point.

 At $\displaystyle (-1.6,-1.786)$,

$\displaystyle D(x,y)=(12(-1.6)-4(-1.786))(12(-1.786)^2)-(-4(-1.6)-7)^2$

$\displaystyle D(x,y)=(-19.2-7.144)(38.278)-(6.4-7)^2=-1008.7< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (-1.6,-1.786)$ is a saddle point.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=4x^3+3y$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=4y^3+3x$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=12x^2$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=12y^2$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=3$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=4x^3+3y=0\Rightarrow 4x^3=-3y \Rightarrow y=\frac{-4x^3}{3}$

$\displaystyle f_y(x,y)=4y^3+3x=0\Rightarrow 4\left ( \frac{-4x^3}{3} \right )^3+3x=0$

$\displaystyle 4\left ( \frac{-64x^9}{27} \right )+3x=0\Rightarrow \frac{-256}{27}x^9+3x=0$

$\displaystyle x\left (\frac{-256}{27}x^8+3 \right )=0$

Setting each factor in the expression equal to $\displaystyle 0$ gives us

$\displaystyle x=0$ and $\displaystyle \frac{-256}{27}x^8+3=0\Rightarrow\frac{-256}{27}x^8=-3$

$\displaystyle x=\sqrt[8]{-3*\frac{27}{-257}}=\pm \sqrt3/2$

The real values of $\displaystyle x$ are $\displaystyle x=0$, $\displaystyle x=-\sqrt3/2$ and $\displaystyle x=\sqrt3/2$

We find the corresponding value of $\displaystyle y$ using $\displaystyle y=\frac{-4x^3}{3}$ (found by rearranging the first derivative)

$\displaystyle y(0)=\frac{-4(0)^3}{3}=0$

$\displaystyle y(-\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=\sqrt3/2$

$\displaystyle y(\sqrt3/2)=\frac{-4(-\sqrt3/2)^3}{3}=-\sqrt3/2$

There are critical points at $\displaystyle (0,0)$,$\displaystyle (-\sqrt3/2,\sqrt3/2)$  and $\displaystyle (\sqrt3/2,-\sqrt3/2)$.  We need to determine if the critical points are maxima or minima using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(12x^2)(12y^2)-(3)^2=144x^2y^2-9$

 At $\displaystyle (0,0)$,

$\displaystyle D(x,y)=144(0)^2(0)^2-9=-9< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (0,0)$ is a saddle point.

 At $\displaystyle (-\sqrt3/2,\sqrt3/2)$,

$\displaystyle D(x,y)=144\left ( \frac{-\sqrt3}{2} \right )^2\left ( \frac{\sqrt3}{2} \right )^2-9$

$\displaystyle D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$\displaystyle f_{xx}(x,y)=12x^2=12\left ( \frac{-\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, $\displaystyle (-\sqrt3/2,\sqrt3/2)$ is a minimum.

At $\displaystyle (\sqrt3/2,-\sqrt3/2)$,

$\displaystyle D(x,y)=144\left ( \frac{\sqrt3}{2} \right )^2\left ( \frac{-\sqrt3}{2} \right )^2-9$

$\displaystyle D(x,y)=144\left ( \frac{3}{4} \right )\left ( \frac{3}{4} \right )-9=72>0$

$\displaystyle f_{xx}(x,y)=12x^2=12\left ( \frac{\sqrt3}{2} \right )^2=12\left ( \frac{3}{4} \right )=9>0$

Since $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, $\displaystyle (-\sqrt3/2,\sqrt3/2)$ is a minimum.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 $\displaystyle f(x,y)=sin(x)-cos(y)+4xy$

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=cos(x)+4y$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=sin(y)+4x$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=-sin(x)$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=cos(y)$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=4$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=cos(x)+4y=0\Rightarrow cos(x)=-4y$

$\displaystyle y=-cos(x)/4$

$\displaystyle f_y(x,y)=sin(y)+4x=0\Rightarrow sin\left (\frac{-cos(x)}{4} \right )+4x=0$

Using a TI-83 or other software to find the root, we find that  $\displaystyle x\approx 0.0617$, 

We find the corresponding value of $\displaystyle y$ using $\displaystyle y=-cos(x)/4$ (found by rearranging the first derivative)

$\displaystyle y(0.0617)=-cos(0.0617)/4 \approx-0.2495$

There is a critical points at $\displaystyle (0.0617,-0.2495)$.  We need to determine if the critical point is a maximum or minimum using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=-sin(x)cos(y)-4^2$

 At $\displaystyle (0.0617,-0.2495)$,

$\displaystyle D(x,y)=-sin(0.0617)cos(-0.2495)-4^2\approx -16.06< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (0.0617,-0.2495)$ is a saddle point.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2sin(x)cos(x)$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=2cos(y)(-sin(y))=-2sin(y)cos(y)$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2\left [ sin(x)(-sin(x))+cos(x)cos(x)\right ]$

$\displaystyle f_{xx}(x,y)=2\left [ -sin^2(x)+cos^2(x)\right ]=2cos^2(x)-2sin^2(x)$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=-2\left [ sin(y)(-sin(y))+cos(y)cos(y)\right ]$

$\displaystyle f_{yy}(x,y)=-2\left [ -sin^2(y)+cos^2(y)\right ]=2sin^2(y)-2cos^2(y)$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=2sin(x)cos(x)=0$

$\displaystyle sin(x)=0\Rightarrow x=0, \pi$

$\displaystyle cos(x)=0\Rightarrow x= \pi/2, 3\pi/2$

$\displaystyle f_y(x,y)=-2sin(y)cos(y)=0$

$\displaystyle sin(y)=0\Rightarrow y=0, \pi$

$\displaystyle cos(y)=0\Rightarrow y= \pi/2, 3\pi/2$

Our derivatives equal $\displaystyle 0$ when $\displaystyle x=0, \pi/2, \pi, 3\pi/2$ and $\displaystyle y=0, \pi/2, \pi, 3\pi/2$.  Every linear combination of these points is a critical point.  The critical points are

 $\displaystyle (0,0)$, $\displaystyle (0,\pi/2)$, $\displaystyle (0,\pi)$, $\displaystyle (0,3\pi/2)$

 $\displaystyle (\pi/2,0)$, $\displaystyle (\pi/2,\pi/2)$, $\displaystyle (\pi/2,\pi)$, $\displaystyle (\pi/2,3\pi/2)$

 $\displaystyle (\pi,0)$, $\displaystyle (\pi,\pi/2)$, $\displaystyle (\pi,\pi)$, $\displaystyle (\pi,3\pi/2)$

 $\displaystyle (3\pi/2,0)$, $\displaystyle (3\pi/2,\pi/2)$, $\displaystyle (3\pi/2,\pi)$, $\displaystyle (3\pi/2,3\pi/2)$

We need to determine if the critical point is a maximum or minimum using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(2cos^2x-2sin^2x)(2sin^2y-2cos^2y)-0^2$

$\displaystyle f_{xx}(x,y)=2cos^2(x)-2sin^2(x)$

 $\displaystyle (0,0)$, $\displaystyle (0,\pi/2)$, $\displaystyle (0,\pi)$, $\displaystyle (0,3\pi/2)$

$\displaystyle D(0,0)=(2cos^2(0)-2sin^2(0))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(1)-0)(2(0)-2(1))=(2)(-2)=-4< 0$

Saddle point

$\displaystyle D(0,\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(1)-2(0))(2(1)-2(0))=(2)(2)=4>0$

$\displaystyle f_{xx}(0,\pi/2)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$\displaystyle D(0,\pi)=(2cos^2(0)-2sin^2(0))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(1)-0)(2(0)-2(-1))=(2)(2)=4>0$

$\displaystyle f_{xx}(0,\pi)=2cos^2(0)-2sin^2(0)=2(1)-2(0)=2>0$

minimum

$\displaystyle D(0,3\pi/2)=(2cos^2(0)-2sin^2(0))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(1)-2(0))(2(-1)-2(0))=(2)(-2)=-4< 0$

Saddle point

 $\displaystyle (\pi/2,0)$, $\displaystyle (\pi/2,\pi/2)$, $\displaystyle (\pi/2,\pi)$, $\displaystyle (\pi/2,3\pi/2)$

$\displaystyle D(\pi/2,0)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(0)-2(1))(2(0)-2(1))=(-2)(-2)=4>0$

$\displaystyle f_{xx}(\pi/2,0)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2< 0$

maximum

$\displaystyle D(\pi/2,\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(0)-2(1))(2(1)-2(0))=(-2)(2)=-4< 0$

saddle point

$\displaystyle D(\pi/2,\pi)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(0)-2(1))(2(0)-2(-1))=(-2)(2)=-4< 0$

saddle point

$\displaystyle D(\pi/2,3\pi/2)=(2cos^2(\pi/2)-2sin^2(\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(0)-2(1))(2(-1)-2(0))=(-2)(-2)=4>0$

$\displaystyle f_{xx}(\pi/2,2\pi/2)=2cos^2(\pi/2)-2sin^2(\pi/2)=2(0)-2(1)=-2< 0$

maximum

 $\displaystyle (\pi,0)$, $\displaystyle (\pi,\pi/2)$, $\displaystyle (\pi,\pi)$, $\displaystyle (\pi,3\pi/2)$

$\displaystyle D(\pi,0)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(-1)-0)(2(0)-2(1))=(-2)(-2)=4>0$

$\displaystyle f_{xx}(\pi,\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2< 0$

maximum

$\displaystyle D(\pi,\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(-1)-2(0))(2(1)-2(0))=(-2)(2)=-4< 0$

saddle point

$\displaystyle D(\pi,\pi)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(-1)-0)(2(0)-2(-1))=(-2)(2)=-4< 0$

saddle point

$\displaystyle D(\pi,3\pi/2)=(2cos^2(\pi)-2sin^2(\pi))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(-1)-2(0))(2(-1)-2(0))=(-2)(-2)=4>0$

$\displaystyle f_{xx}(\pi,3\pi/2)=2cos^2(\pi)-2sin^2(\pi)=2(-1)-2(0)=-2< 0$

maximum

 $\displaystyle (3\pi/2,0)$, $\displaystyle (3\pi/2,\pi/2)$, $\displaystyle (3\pi/2,\pi)$, $\displaystyle (3\pi/2,3\pi/2)$

$\displaystyle D(3\pi/2,0)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(0)-2cos^2(0))-0^2$

$\displaystyle =(2(0)-2(-1))(2(0)-2(1))=(2)(-2)=-4< 0$

saddle point

$\displaystyle D(3\pi/2,\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi/2)-2cos^2(\pi/2))-0^2$

$\displaystyle =(2(0)-2(-1))(2(1)-2(0))=(2)(2)=4>0$

$\displaystyle f_{xx}(3\pi/2,\pi/2)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$\displaystyle D(3\pi/2,\pi)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(\pi)-2cos^2(\pi))-0^2$

$\displaystyle =(2(0)-2(-1))(2(0)-2(-1))=(2)(2)=4>0$

$\displaystyle f_{xx}(3\pi/2,\pi)=2cos^2(3\pi/2)-2sin^2(3\pi/2)=2(0)-2(-1)=2>0$

minimum

$\displaystyle D(3\pi/2,3\pi/2)=(2cos^2(3\pi/2)-2sin^2(3\pi/2))(2sin^2(3\pi/2)-2cos^2(3\pi/2))-0^2$

$\displaystyle =(2(0)-2(-1))(2(-1)-2(0))=(2)(-2)=-4< 0$

saddle point

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=2x$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=2y$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=2$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=2$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=0$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=2x=0\Rightarrow x=0$

$\displaystyle f_y(x,y)=2y=0\Rightarrow y=0$

There is a critical point at $\displaystyle (0,0)$.  We need to determine if the critical point is a maximum or minimum using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=2(2)-0^2=4$

 At $\displaystyle (0,0)$,

$\displaystyle D(x,y)=4>0$

$\displaystyle f_{xx}(x,y)=2>0$

Since $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at $\displaystyle (0,0)$.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

$\displaystyle f(x,y)=3x^2y^2-3x^2$

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=6xy^2-6x$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=6x^2y$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=6y^2-6$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=6x^2$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=12xy$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_{xx}(x,y)=6y^2-6=0\Rightarrow 6y^2=6\Rightarrow y=\pm1$

$\displaystyle f_{yy}(x,y)=6x^2=0 \Rightarrow x=0$

The critical points are  $\displaystyle (-1,0)$ and $\displaystyle (1,0)$.  We need to determine if the critical point is a maximum or minimum using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(6y^2-6)(6x^2)-(12xy)^2=36x^2y^2-36x^2-144x^2y^2$

$\displaystyle D(x,y)=-108x^2y^2-36x^2$

 At $\displaystyle (-1,0)$,

$\displaystyle D(x,y)=-108(-1)^2(0)^2-36(-1)^2=-36< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (-1,0)$ is a saddle point.

At $\displaystyle (1,0)$,

$\displaystyle D(x,y)=-108(-1)^2(0)^2-36(1)^2=-36< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (1,0)$ is a saddle point.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^3+5y^2-2x}(15x^2-2)=(15x^2-2)e^{5x^3+5y^2-2x}$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^3+5y^2-2x}(10y)=(10y)e^{5x^3+5y^2-2x}$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}(15x^2-2)+e^{5x^3+5y^2-2x}(30x)$

$\displaystyle f_{xx}(x,y)=((15x^2-2)^2+30x)e^{5x^3+5y^2-2x}$

$\displaystyle f_{xx}(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=(10y)e^{5x^3+5y^2-2x}(10y)+e^{5x^3+5y^2-2x}(10)$

$\displaystyle f_{yy}(x,y)=(100y^2)e^{5x^3+5y^2-2x}+(10)e^{5x^3+5y^2-2x}$

$\displaystyle f_{yy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=(100y^2+10)e^{5x^3+5y^2-2x}(15x^2-2)$

$\displaystyle f_{xy}(x,y)=(100y^2+10)(15x^2-2)e^{5x^3+5y^2-2x}$

$\displaystyle f_{xy}(x,y)=(1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x}$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle f_x(x,y)=(15x^2-2)e^{5x^3+5y^2-2x}$

$\displaystyle f_y(x,y)=(10y)e^{5x^3+5y^2-2x}$

The exponential part of each expression cannot equal $\displaystyle 0$, so each derivative is $\displaystyle 0$ only when $\displaystyle (15x^2-2)=0$ and $\displaystyle 10y=0$.  That is $\displaystyle x=\pm \sqrt{2/15}$ and $\displaystyle y=0$.

The critical points are $\displaystyle (-\sqrt{2/15},0)$ and $\displaystyle (\sqrt{2/15},0)$.  We need to determine if the critical points are maxima or minima using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(225x^4-60x^2+30x+4)e^{5x^3+5y^2-2x}(100y^2+10)e^{5x^3+5y^2-2x} -((1500x^2y^2+150x^2-200y^2-20)e^{5x^3+5y^2-2x})^2$

$\displaystyle D(x,y)=(225x^4-60x^2+30x+4)(100y^2+10)(e^{5x^3+5y^2-2x})^2 -(1500x^2y^2+150x^2-200y^2-20)^2(e^{5x^3+5y^2-2x})^2$

 At $\displaystyle (5/4,0)$,

$\displaystyle D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(0+10)(e^{5(5/4)^3+5(0)^2-2(5/4)})^2 -(1500(5/4)^2(0)^2+150(5/4)^2-200(0)^2-20)^2(e^{5(5/4)^3+5(0)^2-2(5/4)})^2$

$\displaystyle D(x,y)=(225(5/4)^4-60(5/4)^2+30(5/4)+4)(10)(e^{5(5/4)^3-2(5/4)})^2 -(150(5/4)^2-20)^2(e^{5(5/4)^3-2(5/4)})^2$

$\displaystyle D(x,y)\approx(497)(10)(2.0457) -(45956.6)(2.0457)$

$\displaystyle D(x,y)\approx -83846.3< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (5/4,0)$ is a saddle point.

To find the relative maxima and minima, we must find all the first order and second order partial derivatives.  We will use the first order partial derivative to find the critical points, then use the equation

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

to classify the critical points.  

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)>0$, then there is a relative minimum at this point.

If $\displaystyle D(x,y)>0$ and $\displaystyle f_{xx}(x,y)< 0$, then there is a relative maximum at this point.

If $\displaystyle D(x,y)< 0$, then this point is a saddle point.

If $\displaystyle D(x,y)=0$, then this point cannot be classified.

 The first order partial derivatives are

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The second order partial derivatives are

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial x} \right )=f_{xx}(x,y)=e^{5x^2-7xy}(10)+(10x-7y)e^{5x^2-7xy}(10x-7y)$

$\displaystyle f_{xx}(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}$

$\displaystyle \frac{\partial }{\partial y} \left (\frac{\partial f }{\partial y} \right )=f_{yy}(x,y)=e^{5x^2-7xy}(-7x)(-7x)$

$\displaystyle f_{yy}(x,y)=90x^2*e^{5x^2-7xy}$

$\displaystyle \frac{\partial }{\partial x} \left (\frac{\partial f }{\partial y} \right )=\frac{\partial }{\partial y} \left (\frac{\partial f }{\partial x} \right )=f_{xy}(x,y)=e^{5x^2-7xy}(-7)+(-7x)e^{5x^2-7xy}(10x-7y)$

$\displaystyle f_{xy}(x,y)=(-7+(-7x)(10x-7y))e^{5x^2-7xy}$

$\displaystyle f_{xy}(x,y)=(-7-70x^2+49xy)e^{5x^2-7xy}$

To find the critical points, we will set the first derivatives equal to $\displaystyle 0$

$\displaystyle \frac{\partial f }{\partial x}=f_x(x,y)=e^{5x^2-7xy}(10x-7y)$

$\displaystyle \frac{\partial f }{\partial y}=f_y(x,y)=e^{5x^2-7xy}(-7x)$

The exponential part of each expression cannot equal $\displaystyle 0$, so each derivative is $\displaystyle 0$ only when $\displaystyle (10x-7y)=0$ and $\displaystyle -7x=0$.  That is $\displaystyle x=0$ and $\displaystyle y=0$.

The critical point is $\displaystyle (0,0)$.  We need to determine if the critical points are maxima or minima using $\displaystyle D(x,y)$ and $\displaystyle f_{xx}(x,y)$.  

$\displaystyle D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left ( f_{xy}(x,y) \right )^2$

$\displaystyle D(x,y)=(10+(10x-7y)^2)e^{5x^2-7xy}*90x^2*e^{5x^2-7xy}-((-7-70x^2+49xy)e^{5x^2-7xy})^2$

$\displaystyle D(x,y)=(90x^2)(10+(10x-7y)^2)(e^{5x^2-7xy})^2 -(-7-70x^2+49xy)^2(e^{5x^2-7xy})^2$

$\displaystyle D(x,y)=\left [(90x^2)(10+(10x-7y)^2)-(-7-70x^2+49xy)^2 \right ](e^{5x^2-7xy})^2$

$\displaystyle D(x,y)=(4100x^4-5740x^3y+2009x^2y^2-80x^2+686xy-49)(e^{5x^2-7xy})^2$

 At $\displaystyle (0,0)$,

$\displaystyle D(x,y)=(4100(0)-5740(0)+2009(0)-80(0)+686(0)-49)(e^{5(0)-7(0)})^2$

$\displaystyle D(x,y)=(-49)(1)^2< 0$

Since $\displaystyle D(x,y)< 0$, $\displaystyle (0,0)$ is a saddle point.

The first thing we have to do is figure out the general equations for the lines that create the trapezoid.

$\displaystyle y=x$

$\displaystyle y=-x$

$\displaystyle y=-x+5$

$\displaystyle y=x-5$

Now we have the general equations for out trapezoid, now we need to plug in our transformations into these equations.

$\displaystyle y=x$

$\displaystyle 2u+3v=2u-3v$

$\displaystyle -6v=0\rightarrow v=0$

$\displaystyle y=-x$

$\displaystyle 2u+3v=-2u+3v$

$\displaystyle 4u=0\rightarrow u=0$

$\displaystyle y=-x+5$

$\displaystyle 2u+3v=-2u+3v+5$

$\displaystyle 4u=5\rightarrow u=\frac{5}{4}$

$\displaystyle y=x-5$

$\displaystyle 2u+3v=2u-3v-5$

$\displaystyle -6v=-5\rightarrow v=\frac{5}{6}$

So our region is a rectangle given by $\displaystyle 0 \leq u \leq \frac{5}{4}$, $\displaystyle 0 \leq v \leq \frac{5}{6}$

Next we need to calculate the Jacobian.

$\displaystyle \begin{vmatrix} 2 & 3\\ 2 & -3 \end{vmatrix} =-6-6=-12$

Now we can put the integral together.

$\displaystyle \int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} (3(2u+3v)+2(2u-3v))) \cdot \left | -12\right | \ du\ dv$

$\displaystyle =\int_{0}^{\frac{5}{6}} \int_{0}^{\frac{5}{4}} 120u+36v \ du\ dv$

$\displaystyle =\int_{0}^{\frac{5}{6}} 60u^2+36uv \Big|_{0}^{\frac{5}{4}}\ dv$

$\displaystyle =\int_{0}^{\frac{5}{6}} 60( \frac{5}{4})^2+36(\frac{5}{4})v-0 \ dv$

$\displaystyle =\int_{0}^{\frac{5}{6}} \frac{375}{4}+45v \ dv$

$\displaystyle =\frac{375}{4}v+\frac{45}{2}v^2\Big|_{0}^{\frac{5}{6}} \ dv$

$\displaystyle =\frac{375}{4}\cdot\frac{5}{6}+\frac{45}{2}\cdot\Big(\frac{5}{6}\Big)^2 -0=\frac{375}{4}$