Consider the function f\left ( x, \ y \right ) = ax^2 + by^2. a) Show that (0, 0) is a critical...

Question:

Consider the function {eq}f\left ( x, \ y \right ) = ax^2 + by^2 {/eq}.

a) Show that (0, 0) is a critical point.

b) Compute Hessian H(x, y) and determine D(x, y).

c) Find conditions on a and b so that

i) (0, 0) is local min.

ii) (0, 0) is local max.

iii) (0, 0) is local saddle point.

Finding critical points with the second derivative test

It is possible that functions with two independent variables have constants, then these constants can exist at the first and the second partial derivatives, so to make the second derivative test we need to assign values to the constant. Remember:

Classification of Critical Points (Hessian, or second derivative test):

Suppose the point {eq}(a,b) {/eq} is said to be a critical point of {eq}f(x,y) {/eq} and that the second order partial derivatives are continuous in some region that contains {eq}(a,b) {/eq}.

Now the hessian is, or the second derivative test, is:

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

Then We will have the following classifications for the critical point.

1. If {eq}D>0 {/eq} and {eq}f_{xx}(a,b)>0 {/eq} then there is a local minimum at {eq}(a,b) {/eq}.

2.If {eq}D>0 {/eq} and {eq}f_{xx}(a,b)<0 {/eq} then there is a local maximum at {eq}(a,b) {/eq}.

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

4. If {eq}D=0 {/eq} then we can't say anything about the point {eq}(a,b) {/eq} or the test is inconclusive.

The function is

{eq}f(x,y)=ax^2 + by^2 {/eq}

The first partial derivatives are:

{eq}\displaystyle f_x= 2\,ax \\ \displaystyle f_y= 2\,by \\ {/eq}

a)

{eq}(0, 0) {/eq} is a critical point because, {eq}\displaystyle f_x=0 \; \Rightarrow \; 0= 2\,ax \; \Leftrightarrow \; x=0 \; \text{or} \;a=0 \\ \displaystyle f_y=0 \; \Rightarrow \; 0= 2\,by \; \Leftrightarrow \; y=0 \; \text{or} \;b=0 \\ {/eq}

The second partial derivatives are:

{eq}\displaystyle f_{xx}=2\,a\\ \displaystyle f_{yy}= 2b \\ \displaystyle f_{xy}=f_{yx}=0 \\ {/eq}

b)

The hessian is:

{eq}\displaystyle H(x,y)=f_{xx}(a,b) f_{yy}(a,b)-[f_{xy}(a,b)]^2 {/eq}

Therefore, the second derivative test, for the point {eq}(0,0) {/eq} is

{eq}\displaystyle D(0,0)= f_{xx}(0,0)*f_{yy}-[f_{xy}(0,0)]^2 \; \Rightarrow \; D(0,0)= 2a*2b-[0]^2 \; \Rightarrow \; D(0,0)= 2a*2b \\ \displaystyle D(0,0)= 4ab \\ {/eq}

c)Therefore:

i) (0, 0) is local min if:

{eq}\displaystyle a \; \in \; (0, \infty) {/eq} & {eq}\displaystyle b \; \in \; (0, \infty) {/eq}

ii) (0, 0) is local max if: {eq}\displaystyle a \; \in \; (-\infty, 0) {/eq} & {eq}\displaystyle b \; \in \; (0, \infty) {/eq}

or

{eq}\displaystyle a \; \in \; (0, \infty){/eq} & {eq}\displaystyle b \; \in \; (-\infty, 0) {/eq}

iii) (0, 0) is local saddle point if:

{eq}\displaystyle a = 0 {/eq} or {eq}\displaystyle b = 0 {/eq}