# Find the relative maximum and minimum values and the saddle points, if they exist. f(x,y) = 8xy(x...

## Question:

Find the relative maximum and minimum values and the saddle points, if they exist.

{eq}f(x,y) = 8xy\left(x + y\right) + 6 {/eq}

## Second Partials Test:

A critical point for a function {eq}f(x,y) {/eq} of two variables {eq}x {/eq} and {eq}y {/eq} is a point {eq}(a,b) {/eq} where {eq}f_x(a,b)=f_y(a,b)=0 {/eq} or {eq}f {/eq} is not differentiable at {eq}(a,b). {/eq}

To determine whether a critical point is a local maximum, minimum or saddle point we often use the Second Partials Test which states:

Let {eq}f {/eq} be a function with continuous first and second order derivatives in a disk around {eq}(a,b) {/eq} for which {eq}f_x(a,b)=f_y(a,b)=0. {/eq}

Define the discriminant of {eq}f {/eq} at {eq}(a,b) {/eq} by

{eq}D=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^2 {/eq}

Then

1) If {eq}D>0, {/eq} then {eq}f(a,b) {/eq} is a local extremum. If {eq}f_{xx}(a,b)>0, {/eq} then {eq}f(a,b) {/eq} is a local minimum; If {eq}f_{xx}(a,b)<0, {/eq} then {eq}f(a,b) {/eq} is a local maximum.

2) If {eq}D<0, {/eq} then {eq}(a,b) {/eq} is a saddle point.

3) If {eq}D=0, {/eq} then this test gives no information.

## Answer and Explanation:

Let {eq}f(x,y)=8xy(x+y)+6=8x^2y+8xy^2+6 {/eq}. We then have

{eq}f_x=16xy+8y^2\\ f_y=8x^2+16xy {/eq}

Setting them both equal to 0 (note (0,0) is a solution so we assume {eq}x {/eq} and {eq}y {/eq} are not 0) we have

{eq}16xy+8y^2=0 \text{ and } 8x^2+16xy=0 {/eq}

Or

{eq}8y^2=-16xy \text{ or } y=-2x \text{ and}\\ 8x^2=-16xy \text{ or } x=-2y {/eq}

Thus we have {eq}y=-2x=-2(-2y)=4y {/eq} and this means {eq}y=0 {/eq} and thus {eq}x=0 {/eq}. So the only critical point is {eq}(0,0) {/eq}.

Next we note that

{eq}f_{xx}=16y\\ f_{yy}=16x\\ f_{xy}=16x+16y {/eq}

We can see that the second partials test fails since {eq}D(0,0)=0 {/eq}.

To see this is a saddle point we note that if {eq}y=x {/eq} we have

{eq}f(x,x)=8xx(x+x)+6=16x^3+6 {/eq}. Since 0 is not a local maximum or minimum, we conclude that the {eq}f(x,y) {/eq} has a saddle point at {eq}(0,0) {/eq}.