# Find all critical points for the function and classify each as a relative maximum, a relative...

## Question:

Find all critical points for the function and classify each as a relative maximum, a relative minimum, or a saddle point.

{eq}1. \ f\left ( x, \ y \right ) = x^2y - 2xy^2 + 3xy + 4 \\ 2. \ f\left ( x, \ y \right ) = xy + 8x + 8y \\ 3. \ f\left ( x, \ y \right ) = -x^4 - 32x + y^3 - 12y + 7 \\ 4. \ f\left ( x, \ y \right ) = \left ( x^2 + 2y^2 \right )e^1 - x^2 - y^2 {/eq}

## Hessian

In calculus, polynomail functions with two independent variables have critical points at their first derivatives are zero, it is {eq}\displaystyle f_x(x,y)=0 \; \text{&} \; f_y(x,y)=0 {/eq} Also, we can find critical points ( relative maximum points, relative minimum points, and saddle points) with the hessian or second derivative test. Remember:

Let {eq}f(x,y) {/eq} have a continuous second partial derivatives on an open region containing a point {eq}(a,b) {/eq} for wich: {eq}f_x(a,b)=0 \; \text{&} \; f_y(a,b)=0 {/eq} Then, the test for find relative extrema is:

{eq}\displaystyle D(a,b)= f_{xx}(a,b).f_{yy}(a,b)-[ f_{xy} (a,b) ]^{2} \\ {/eq} If {eq}D(a,b) > 0 \; \text{&} \; f_{xx} > 0 \; \Rightarrow \; \; \text{ f has a relative minimun point at (a,b, f(a,b))} \\ D(a,b) > 0 \; \text{&} \; f_{xx} < 0 \; \Rightarrow \; \; \text{ f has a relative maximun point at (a,b, f(a,b))} \\ D(a,b) < 0 \; \; \Rightarrow \; \; \text{then (a,b, f(a,b)) is a saddle point} \\ D(a,b) = 0 \Rightarrow \; \; \text{ the test is inconclusive} \\ {/eq}

## Answer and Explanation:

(1)

The function is

{eq}\displaystyle \displaystyle f (x, y) = x^2y - 2xy^2 + 3xy + 4 \\ {/eq}

Its first partial derivatives are:

{eq}\displaystyle f_x= 2\,xy-2\,{y}^{2}+3\,y \; \Rightarrow \; f_x= y \left( 2\,x-2\,y+3 \right) \\ \displaystyle f_y={x}^{2}-4\,xy+3\,x \; \Rightarrow \; f_y= x \left( x-4\,y+3 \right) \\ {/eq}

Its second partial derivatives are:

{eq}\displaystyle f_{xx}= 2y\\ \displaystyle f_{yy}= -4x \\ \displaystyle f_{xy}=f_{yx}=2\,x-4\,y+3 \\ {/eq}

We are finding its local extreme values, so

{eq}\displaystyle \begin{align} & f_x(x,y) = 0 \; \Rightarrow \; 0=2\,xy-2\,{y}^{2}+3\,y \; \Rightarrow \; x=y- \frac{3}{2} &(1) \\ \end{align} {/eq}

Similarly,

{eq}\displaystyle \begin{align} & f_y(x,y) = 0 \; \Rightarrow \; 0={x}^{2}-4\,xy+3\,x & \left(\text{using (1)}\right) \\ & 0={(y- \frac{3}{2})}^{2}-4\,(y- \frac{3}{2})y+3\,(y- \frac{3}{2}) \; \Leftrightarrow \; y= \frac{1}{2} \; \text{&} \; y= \frac{3}{2} \\ \end{align} {/eq}

From (1)

{eq}\displaystyle \text{for }y= \frac{1}{2} \Rightarrow x=-1\\ \text{for }y=\frac{3}{2} \Rightarrow x= 0 \\ {/eq}

Therefore the critical points are {eq}\displaystyle (0,0) \; \text{ , } \; ~\left(0,\frac{3}{2}\right) \; ~\left(-3,0\right) \; \text{&} \; ( -1, 1/2) \\ {/eq}.

Now to classify these critical points we will use the hessian or the second derivative test:

For {eq}(0,0) {/eq}

{eq}\displaystyle \begin{align} \displaystyle f_{xx}(0,0) &=2(0) \; \Rightarrow \; f_{xx}(0,0)=0 \\ \displaystyle f_{yy}(0,0) &=-4(0) \; \Rightarrow \; f_{yy}(0,0)=0 \\ \displaystyle f_{xy}=f_{yx}=2\,(0)-4\,(0)+3 \; \Rightarrow \; f_{xy}=f_{yx}=3 \\ \end{align} {/eq}

The second derivative test is:

{eq}\displaystyle D(0,0) = (f_{xx}(0,0))(f_{yy}(0,0))-\left(f_{xy}(0,0)\right)^2 \\ \displaystyle D(0,0)= 0*0-(3)^2 \; \Rightarrow \; D(0,0)= -9 \; < \; 0 {/eq}

Therefore the point {eq}\displaystyle (0,0, f(0,0)) \; \Rightarrow \; (0,0, 4) {/eq} is a saddle point for the given function.

So,

{eq}\begin{array} \; \; \text{(a,b)} \; & { \ f_{xx} (a,b) } & { f_{yy} (a,b) } & f_{xy} (a,b) & D(a,b) & Conclusion & (a,b, f(a,b)) \\ \hline (0,0) & 0 & 0 & 3 & -9 & saddle \; & (0,0, 4) \\ \displaystyle ( 0, \frac{3}{2} ) & 3 & 0 & -3 & -9 & saddle & ( 0 , \frac{3}{2} , 4 ) \\ \displaystyle ( -3, 0 ) & 0 & 12 & -3 & -9 & saddle & ( -3 , 0 , 4 ) \\ \displaystyle ( -1, \frac{1}{2} ) & 1 & 4 & -1 & 3 & \text{ relative minimum point} & ( -3 , 0 , 4 ) \\ \end{array} \\ {/eq}

(2)

The function is

{eq}\displaystyle \displaystyle f (x, y) = xy + 8x + 8y \\ {/eq}

Its first partial derivatives are:

{eq}\displaystyle f_x= y+8 \\ \displaystyle f_y=x+8\\ {/eq}

Its second partial derivatives are:

{eq}\displaystyle f_{xx}= 0\\ \displaystyle f_{yy}= 0 \\ \displaystyle f_{xy}=f_{yx}=1 \\ {/eq}

We are finding its local extreme values, so

{eq}\displaystyle \displaystyle f_x(x,y) = 0 \; \Rightarrow \; 0=y+8 \; \Rightarrow \; y=-8 \\ {/eq}

Similarly,

{eq}\displaystyle f_y(x,y) = 0 \; \Rightarrow \; 0=x+8 \; \Rightarrow \; x=-8 \\ {/eq}

Therefore the critical point is {eq}\displaystyle (-8,-8) \\ {/eq}.

Now to classify this critical point we will use the hessian or the second derivative test:

For {eq}(-8,-8) {/eq}

{eq}\displaystyle \displaystyle f_{xx}(-8,-8) =0\\ \displaystyle f_{yy}(-8,-8) =0\\ \displaystyle f_{xy}=f_{yx}(-8,-8)=1 \\ {/eq}

The second derivative test is:

{eq}\displaystyle D(-8,-8) = (f_{xx}(-8,-8))(f_{yy}(-8,-8))-\left(f_{xy}(-8,-8)\right)^2 \\ \displaystyle D(0,0)= 0*0-(1)^2 \; \Rightarrow \; D(0,0)= -1 \; < \; 0 {/eq}

Therefore the point {eq}\displaystyle (0,0, f(0,0)) \; \Rightarrow \; (0,0, -64) {/eq} is a saddle point for the given function.

(3)

The function is

{eq}\displaystyle \displaystyle f (x, y) = -x^4 - 32x + y^3 - 12y + 7\\ {/eq}

Its first partial derivatives are:

{eq}\displaystyle f_x= -4\,{x}^{3}-32 \; \Rightarrow \; f_x= -4\, \left( x+2 \right) \left( {x}^{2}-2\,x+4 \right)\\ \displaystyle f_y=3\,{y}^{2}-12 \; \Rightarrow \; f_y=3\, \left( y-2 \right) \left( y+2 \right) \\ {/eq}

Its second partial derivatives are:

{eq}\displaystyle f_{xx}= -12\,{x}^{2} \\ \displaystyle f_{yy}= 6\,y \\ \displaystyle f_{xy}=f_{yx}= 0 \\ {/eq}

We are finding its local extreme values, so

{eq}\displaystyle \displaystyle f_x(x,y) = 0 \; \Rightarrow \; 0=-4\, \left( x+2 \right) \left( {x}^{2}-2\,x+4 \right) \; \Leftrightarrow \; x=-2 \\ {/eq}

Similarly,

{eq}\displaystyle \displaystyle f_y(x,y) = 0 \; \Rightarrow \; 0=3\, \left( y-2 \right) \left( y+2 \right) \; \Leftrightarrow \; y= -2 \; \text{&} \; y=2 \\ {/eq}

Therefore the critical points are {eq}\displaystyle (-2,2) \; \text{&} \; ~\left(-2,-2\right) \\ {/eq}.

Now to classify these critical points we will use the hessian or the second derivative test:

For {eq}(-2,2) {/eq}

{eq}\displaystyle \begin{align} \displaystyle f_{xx}(-2,2) &=-12\,{(-2)}^{2} \; \Rightarrow \; f_{xx}(-2,2)=-48 \\ \displaystyle f_{yy}(-2,2) &=6(2) \; \Rightarrow \; f_{yy}(-2,2)=12 \\ \displaystyle f_{xy}=f_{yx}(-2,2)=0 \\ \end{align} {/eq}

The second derivative test is:

{eq}\displaystyle D(-2,2) = (f_{xx}(-2,2))(f_{yy}(-2,2))-\left(f_{xy}(-2,2)\right)^2 \\ \displaystyle D(-2,2)= -48*12-(0)^2 \; \Rightarrow \; D(-2,2)= -576 \; < \; 0 {/eq}

Therefore the point {eq}\displaystyle (-2,2, f(-2,2)) \; \Rightarrow \; (-2,2, 39) {/eq} is a saddle point for the given function.

So,

{eq}\begin{array} \; \; \text{(a,b)} \; & { \ f_{xx} (a,b) } & { f_{yy} (a,b) } & f_{xy} (a,b) & D(a,b) & Conclusion & (a,b, f(a,b)) \\ \hline (-2,2) & -48 & 12 & 0 & -576 & saddle \; & (0,0, 39) \\ \displaystyle ( -2, -2 ) & -48 & -12 & 0 & 576 & relative \; maximum \; & (-2,-2, 71) \\ \end{array} \\ {/eq}

(4)

The function is

{eq}\displaystyle \displaystyle f (x, y) =\left ( x^2 + 2y^2 \right )e^1 - x^2 - y^2 \\ {/eq}

Its first partial derivatives are:

{eq}\displaystyle f_x= 2\,x{\rm e}-2\,x \; \Rightarrow \; f_x= 2\,x \left( {\rm e}-1 \right) \\ \displaystyle f_y= 4\,y{\rm e}-2\,y \; \Rightarrow \; f_y=2\,y \left( 2\,{\rm e}-1 \right) \\ {/eq}

Its second partial derivatives are:

{eq}\displaystyle f_{xx}= 2\,{\rm e}-2 \\ \displaystyle f_{yy}=4\,{\rm e}-2\\ \displaystyle f_{xy}=f_{yx}=0\\ {/eq}

We are finding its local extreme values, so

{eq}\displaystyle \displaystyle f_x(x,y) = 0 \; \Rightarrow \; 0= 2\,x \left( {\rm e}-1 \right) \; \Leftrightarrow \; x=0 \\ {/eq}

Similarly,

{eq}\displaystyle \displaystyle f_y(x,y) = 0 \; \Rightarrow \; 0= 2\,y \left( 2\,{\rm e}-1 \right) \; \Leftrightarrow \; y=0 \\ {/eq}

Therefore the critical point is: {eq}\displaystyle (0,0) \\ {/eq}.

Now to classify these critical points we will use the hessian or the second derivative test:

For {eq}(0,0) {/eq}

{eq}\displaystyle \begin{align} \displaystyle f_{xx} (0,0) &=2\,{\rm e}-2 \; \Rightarrow \; f_{xx} (0,0)=3.4365636569181 \\ \displaystyle f_{yy} (0,0) &=4\,{\rm e}-2 \; \Rightarrow \; f_{yy} (0,0)= 8.8731273138362\\ \displaystyle f_{xy}=f_{yx} (0,0)=0 \\ \end{align} {/eq}

The second derivative test is:

{eq}\displaystyle D(0,0)= (f_{xx}(0,0))(f_{yy}(0,0))-\left(f_{xy}(0,0)\right)^2 \\ \displaystyle D(-2,2)= (2\,{\rm e}-2)*(4\,{\rm e}-2)-(0)^2 \; \Rightarrow \; D(0,0)= 30.4930668499367 \; > \; 0 {/eq}

Therefore the point {eq}\displaystyle (0,0, f(0,0)) \; \Rightarrow \; (0,0, 0) {/eq} is a relative minimum point for the given function.